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网络流专题
阅读量：4984 次
发布时间：2019-06-12

本文共 42144 字，大约阅读时间需要 140 分钟。

 

POJ 1149 PIGS 

#include
      
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               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 10005, INF = 1e9;struct Edge{ int to, nxt, cap;}e[N << 1];int head[N], cur[N], dis[N], q[N], En = 1, S, T;void add_edge(int u,int v,int w) {// cout << u << " " << v << " " << w << "\n"; ++En; e[En].to = v; e[En].nxt = head[u]; e[En].cap = w; head[u] = En; ++En; e[En].to = u; e[En].nxt = head[v]; e[En].cap = 0; head[v] = En;}bool bfs() { for (int i = 0; i <= T; ++i) dis[i] = -1, cur[i] = head[i]; int L = 1, R = 0; q[++R] = S; dis[S] = 0; while (L <= R) { int u = q[L ++]; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == -1 && e[i].cap > 0) { dis[v] = dis[u] + 1; q[++R] = v; if (v == T) return 1; } } } return 0;}int dfs(int u,int flow) { if (u == T) return flow; int used = 0, t; for (int &i = cur[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == dis[u] + 1 && e[i].cap > 0) { t = dfs(v, min(e[i].cap, flow - used)); if (t > 0) { used += t, e[i].cap -= t, e[i ^ 1].cap += t; if (used == flow) break; } } } if (used != flow) dis[u] = -1; return used; }int dinic() { int ans = 0; while (bfs()) ans += dfs(S, INF); return ans;}vector
                
                  vec[N];int w[N], cnt[N];int main() { //freopen("1.txt", "r", stdin); int m = read(), n = read(); for (int i = 1; i <= m; ++i) w[i] = read(); for (int i = 1; i <= n; ++i) { int t = read(); while (t--) { int u = read(); vec[u].push_back(i); } cnt[i] = read(); } S = 0; T = n + 1; for (int i = 1; i <= n; ++i) add_edge(i, T, cnt[i]); for (int i = 1; i <= m; ++i) { if (vec[i].size() == 0) continue; add_edge(S, vec[i][0], w[i]); for (int j = 1; j < vec[i].size(); ++j) { add_edge(vec[i][j - 1], vec[i][j], INF); } } cout << dinic(); return 0;}
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POJ 2699 The Maximum Number of Strong Kings

分析：枚举strong king的个数，然后网络流判断是否可行。

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               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 2005, INF = 1e9;struct Edge{ int to, nxt, cap;}e[N << 1];int head[N], cur[N], dis[N], q[N], En = 1, S, T;void add_edge(int u,int v,int w) {// cout << u << " " << v << " " << w << "\n"; ++En; e[En].to = v; e[En].nxt = head[u]; e[En].cap = w; head[u] = En; ++En; e[En].to = u; e[En].nxt = head[v]; e[En].cap = 0; head[v] = En;}bool bfs() { for (int i = 0; i <= T; ++i) dis[i] = -1, cur[i] = head[i]; int L = 1, R = 0; q[++R] = S; dis[S] = 0; while (L <= R) { int u = q[L ++]; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == -1 && e[i].cap > 0) { dis[v] = dis[u] + 1; q[++R] = v; if (v == T) return 1; } } } return 0;}int dfs(int u,int flow) { if (u == T) return flow; int used = 0, t; for (int &i = cur[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == dis[u] + 1 && e[i].cap > 0) { t = dfs(v, min(e[i].cap, flow - used)); if (t > 0) { used += t, e[i].cap -= t, e[i ^ 1].cap += t; if (used == flow) break; } } } if (used != flow) dis[u] = -1; return used; }int dinic() { int ans = 0; while (bfs()) ans += dfs(S, INF); return ans;}int n, tot, a[N], id[N][N];bool solve(int k) { En = 1; memset(head, 0, sizeof(head)); S = 0; T = n + tot + 1; for (int i = 1; i <= n; ++i) add_edge(S, i, a[i]); for (int i = n + 1; i <= n + tot; ++i) add_edge(i, T, 1); for (int i = 1; i <= n; ++i) { for (int j = i + 1; j <= n; ++j) { if (i >= n - k + 1 && a[i] < a[j]) add_edge(i, id[i][j] + n, 1); else add_edge(i, id[i][j] + n, 1), add_edge(j, id[i][j] + n, 1); } } return dinic() == tot;}void solve() { n = tot = 0; string s; getline(cin, s); int now = 0;// for (int i = 0; i < (int)s.size(); ++i) {// if (s[i] == ' ') a[++n] = now, now = 0;// else now = now * 10 + s[i] - '0';// } // a[++n] = now; for (int i = 0; i < (int)s.size(); ++i) { // 注意此处的读入！！！不要向上面一样读入！！！ if (isdigit(s[i])) { now = 0; while (i < s.size() && isdigit(s[i])) { now = now * 10 + (s[i] - '0'); i ++; } i --; a[++n] = now; } } for (int i = 1; i <= n; ++i) for (int j = i + 1; j <= n; ++j) id[i][j] = id[j][i] = ++tot; for (int i = n; i >= 1; --i) if (solve(i)) { printf("%d\n", i); return ; } puts("0");}int main() { int T = read(); for (; T--; ) solve(); return 0;}
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ZOJ 2760 How Many Shortest Path 

分析：求出那些边可以存在于最短路上，然后跑网络流。

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               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 2005, INF = 1e9;struct Edge{ int to, nxt, cap;}e[30005];int head[N], cur[N], dis[N], q[N], En = 1, S, T, n;void add_edge(int u,int v,int w) {// printf("%d %d %d\n", u, v, w); ++En; e[En].to = v; e[En].nxt = head[u]; e[En].cap = w; head[u] = En; ++En; e[En].to = u; e[En].nxt = head[v]; e[En].cap = 0; head[v] = En;}bool bfs() { for (int i = 1; i <= n; ++i) dis[i] = -1, cur[i] = head[i]; int L = 1, R = 0; q[++R] = S; dis[S] = 0; while (L <= R) { int u = q[L ++]; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == -1 && e[i].cap > 0) { dis[v] = dis[u] + 1; q[++R] = v; if (v == T) return 1; } } } return 0;}int dfs(int u,int flow) { if (u == T) return flow; int used = 0, t; for (int &i = cur[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == dis[u] + 1 && e[i].cap > 0) { t = dfs(v, min(e[i].cap, flow - used)); if (t > 0) { used += t, e[i].cap -= t, e[i ^ 1].cap += t; if (used == flow) break; } } } if (used != flow) dis[u] = -1; return used; }int dinic() { int ans = 0; while (bfs()) ans += dfs(S, INF); return ans;}int g[N][N], t[N][N]; void solve() { En = 1; memset(head, 0, sizeof(head)); if (S == T) { puts("inf"); return ; } for (int k = 1; k <= n; ++k) for (int i = 1;i <= n; ++i) for (int j = 1; j <= n; ++j) if (g[i][k] != -1 && g[k][j] != -1 && (g[i][k] + g[k][j] < g[i][j] || g[i][j] == -1)) g[i][j] = g[i][k] + g[k][j]; if (g[S][T] == -1) { puts("0"); return ; } for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) if (i != j && g[S][T] == g[S][i] + t[i][j] + g[j][T] && g[S][i] != -1 && g[j][T] != -1) add_edge(i, j, 1); printf("%d\n", dinic()); }int main() {// freopen("1.txt", "r", stdin); while (~scanf("%d", &n)) { for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) t[i][j] = g[i][j] = read(); for (int i = 1; i <= n; ++i) t[i][i] = g[i][i] = 0; S = read() + 1, T = read() + 1; solve(); } return 0;}
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WOJ 124 Football match

题意：

有 N 支球队，互相之间已经进行了一些比赛，还剩下M场没有比。现在给出各 支球队目前的总分以及还剩下哪 M 场没有比，问能否合理安排这 M 场比赛的结 果，使得第 N 支球队最后的总分大于其他任何一支球队的总分。已知每场比赛 胜者得 2 分，败者 0 分，平局则各得 1 分。（1 <= N <= 100, 0 <= M <= 1000）

分析：首先让所有n参见的比赛，让n赢，然后剩下的建图跑最大流，判断是否可行。S向每个比赛连，流量为2，每个比赛向两个球队连，流量为2，每个球队向T连，流量为w[n]-w[i]-1，表示最大的得分。

#include
      
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               #define fi(s) freopen(s,"r",stdin);#define fo(s) freopen(s,"w",stdout);using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 10005, INF = 1e9;struct Edge{ int to, nxt, cap;}e[N << 1];int head[N], cur[N], dis[N], q[N], En = 1, S, T;void add_edge(int u,int v,int w) { // printf("%d %d %d\n", u, v, w); ++En; e[En].to = v; e[En].nxt = head[u]; e[En].cap = w; head[u] = En; ++En; e[En].to = u; e[En].nxt = head[v]; e[En].cap = 0; head[v] = En;}bool bfs() { for (int i = 0; i <= T; ++i) dis[i] = -1, cur[i] = head[i]; int L = 1, R = 0; q[++R] = S; dis[S] = 0; while (L <= R) { int u = q[L ++]; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == -1 && e[i].cap > 0) { dis[v] = dis[u] + 1; q[++R] = v; if (v == T) return 1; } } } return 0;}int dfs(int u,int flow) { if (u == T) return flow; int used = 0, t; for (int &i = cur[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == dis[u] + 1 && e[i].cap > 0) { t = dfs(v, min(e[i].cap, flow - used)); if (t > 0) { used += t, e[i].cap -= t, e[i ^ 1].cap += t; if (used == flow) break; } } } if (used != flow) dis[u] = -1; return used; }int dinic() { int ans = 0; while (bfs()) ans += dfs(S, INF); return ans;}int n, m, a[N], b[N], w[N];void solve() { En = 1; memset(head, 0, sizeof(head)); int tot = 0, now = 0; S = 0; for (int i = 1; i <= m; ++i) { if (a[i] == n || b[i] == n) w[n] += 2, a[i] = b[i] = 0; else { now ++; add_edge(S, now, 2); } } // cout << "match" << now << "\n"; for (int i = 1; i <= n; ++i) if (w[i] > w[n]) { puts("NO"); return ; } tot = now; now = 0; for (int i = 1; i <= m; ++i) { if (a[i] && b[i]) { now ++; add_edge(now, a[i] + tot, 2); add_edge(now, b[i] + tot, 2); } } T = tot + n + 1; for (int i = 1; i < n; ++i) add_edge(i + tot, T, w[n] - w[i] - 1); puts(dinic() == tot * 2 ? "YES" : "NO");}int main() { // freopen("1.txt", "r", stdin); while (~scanf("%d%d", &n, &m)) { for (int i = 1; i <= n; ++i) w[i] = read(); for (int i = 1; i <= m; ++i) a[i] = read(), b[i] = read(); solve(); } return 0;}
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SGU 438. The Glorious Karlutka River =)

题意：一条宽为W的河，河中有N块石头，每块石头的坐标(Xi, Yi)和每一时刻最大承受人数Ci。有M个游客，每次最远跳D米，每跳一次1秒。问能否所有人穿越这条河流，如果能，最少需要多长时间。 N,M<=50

分析：动态流问题，首先可以通过判断是否联通来得知能否过河（或者利用最差清空的时间是n+m来判断）。然后按时间拆点，建图跑网络流。每个石头拆成两个点，然后对于一条边u,v，从上一秒的u到下一秒的v连一条inf的边，每秒源点也要连出边。

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               #define pa pair
                
                 #define mp make_pairusing namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 100005, INF = 1e9;struct Edge{ int to, nxt, cap;}e[N << 1];int head[N], cur[N], dis[N], q[N], En = 1, S, T, Index;void add_edge(int u,int v,int w) {// printf("%d %d %d\n", u, v, w); ++En; e[En].to = v; e[En].nxt = head[u]; e[En].cap = w; head[u] = En; ++En; e[En].to = u; e[En].nxt = head[v]; e[En].cap = 0; head[v] = En;}bool bfs() { for (int i = 0; i <= Index; ++i) dis[i] = -1, cur[i] = head[i]; int L = 1, R = 0; q[++R] = S; dis[S] = 0; while (L <= R) { int u = q[L ++]; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == -1 && e[i].cap > 0) { dis[v] = dis[u] + 1; q[++R] = v; if (v == T) return 1; } } } return 0;}int dfs(int u,int flow) { if (u == T) return flow; int used = 0, t; for (int &i = cur[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == dis[u] + 1 && e[i].cap > 0) { t = dfs(v, min(e[i].cap, flow - used)); if (t > 0) { used += t, e[i].cap -= t, e[i ^ 1].cap += t; if (used == flow) break; } } } if (used != flow) dis[u] = -1; return used; }int dinic() { int ans = 0; while (bfs()) ans += dfs(S, INF); return ans;}int n, m, D, L, x[N], y[N], w[N], fa[N], chu[N], ru[N];vector< pa > g;vector
                 
                   p;int sqr(int x) { return x * x; }int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }bool pd() { for (int i = 1; i <= n + 2; ++i) fa[i] = i; for (int i = 1; i <= n; ++i) { if (!w[i]) continue; if (y[i] <= D) { int r1 = find(n + 1), r2 = find(i); if (r1 != r2) fa[r2] = r1; g.push_back(mp(0, i)); p.push_back(i); } if (y[i] >= L - D) { int r1 = find(n + 2), r2 = find(i); if (r1 != r2) fa[r2] = r1; p.push_back(i); g.push_back(mp(i, n + 1)); } } for (int i = 1; i <= n; ++i) for (int j = i + 1; j <= n; ++j) { if (!w[i] || !w[j] || i == j) continue; if (sqr(x[i] - x[j]) + sqr(y[i] - y[j]) > D * D) continue; int r1 = find(i), r2 = find(j); if (r1 != r2) fa[r2] = r1; p.push_back(i), p.push_back(j); g.push_back(mp(i, j)); } p.push_back(0), p.push_back(n + 1); sort(p.begin(), p.end()); p.erase(unique(p.begin(), p.end()), p.end()); return find(n + 1) != find(n + 2);}int main() { n = read(), m = read(), D = read(), L = read(); if (D >= L) { puts("1"); return 0; } // D >= L not D > L !!!!! for (int i = 1; i <= n; ++i) { x[i] = read(), y[i] = read(), w[i] = read(); } if (pd()) { puts("IMPOSSIBLE"); return 0; } S = 0, T = n + n + 1; Index = T; for (int i = 1; i <= n; ++i) { ru[i] = i, chu[i] = i + n; if (y[i] <= D) add_edge(S, i, INF), add_edge(i, i + n, w[i]); } chu[S] = ru[S] = S, chu[n + 1] = ru[n + 1] = T; w[T] = INF;int flow = 0; for (int Ti = 1; Ti <= n + m; ++Ti) { flow += dinic(); if (flow >= m) { cout << Ti; return 0; } for (int i = 1; i < (int)p.size() - 1; ++i) ru[p[i]] = ++Index; for (int i = 0; i < (int)g.size(); ++i) { int u = g[i].first, v = g[i].second; if (p[v] != S && chu[u] != T) add_edge(chu[u], ru[v], INF); if (p[u] != S && chu[v] != T) add_edge(chu[v], ru[u], INF); } for (int i = 1; i < (int)p.size() - 1; ++i) { chu[p[i]] = ++Index; add_edge(ru[p[i]], chu[p[i]], w[p[i]]); } } return 0;}
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SPOJ NETADMIN - Smart Network Administrator

分析： 二分一个mid，然后给每条边设置容量为mid，从源点向每个想要链接internet的用户连一条容量为1的边，从1向汇点连一条容量为k的边。

#include
      
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              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 100005, INF = 1e9;struct Edge{ int to, nxt, cap;}e[N << 1];int head[N], cur[N], dis[N], q[N], En = 1, S, T;void add_edge(int u,int v,int w) { ++En; e[En].to = v; e[En].nxt = head[u]; e[En].cap = w; head[u] = En; ++En; e[En].to = u; e[En].nxt = head[v]; e[En].cap = 0; head[v] = En;}bool bfs() { for (int i = 0; i <= T; ++i) dis[i] = -1, cur[i] = head[i]; int L = 1, R = 0; q[++R] = S; dis[S] = 0; while (L <= R) { int u = q[L ++]; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == -1 && e[i].cap > 0) { dis[v] = dis[u] + 1; q[++R] = v; if (v == T) return 1; } } } return 0;}int dfs(int u,int flow) { if (u == T) return flow; int used = 0, t; for (int &i = cur[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == dis[u] + 1 && e[i].cap > 0) { t = dfs(v, min(e[i].cap, flow - used)); if (t > 0) { used += t, e[i].cap -= t, e[i ^ 1].cap += t; if (used == flow) break; } } } if (used != flow) dis[u] = -1; return used; }int dinic() { int ans = 0; while (bfs()) ans += dfs(S, INF); return ans;}#define pa pair
                
                 vector< pa > g;int n, m, k;int h[N];bool check(int c) { En = 1; memset(head, 0, sizeof(head)); S = 0, T = n + 1; for (int i = 1; i <= k; ++i) add_edge(S, h[i], 1); add_edge(1, T, k); for (int i = 0; i < (int)g.size(); ++i) { add_edge(g[i].first, g[i].second, c); add_edge(g[i].second, g[i].first, c); } return dinic() == k;}void solve() { g.clear(); n = read(), m = read(), k = read(); for (int i = 1; i <= k; ++i) h[i] = read(); for (int i = 1; i <= m; ++i) g.push_back(pa(read(), read())); int L = 1, R = k, ans = 0; while (L <= R) { int mid = (L + R) >> 1; if (check(mid)) ans = mid, R = mid - 1; else L = mid + 1; } cout << ans << "\n";}int main() { freopen("1.txt", "r", stdin); for (int T = read(); T --; ) solve(); return 0;}
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SPOJ IM - Intergalactic Map

题意：无向图中，能否从1走到2，再从2走到3，每个点只经过一次。

分析：源点向2连，容量为2，1和3向汇点连，容量为1，建图跑网络流，判断容量是否为2。

#include
      
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              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 200005, INF = 1e9;struct Edge{ int to, nxt, cap;}e[N << 1];int head[N], cur[N], dis[N], q[N], En = 1, S, T;void add_edge(int u,int v,int w) { ++En; e[En].to = v; e[En].nxt = head[u]; e[En].cap = w; head[u] = En; ++En; e[En].to = u; e[En].nxt = head[v]; e[En].cap = 0; head[v] = En;}bool bfs() { for (int i = 0; i <= T; ++i) dis[i] = -1, cur[i] = head[i]; int L = 1, R = 0; q[++R] = S; dis[S] = 0; while (L <= R) { int u = q[L ++]; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == -1 && e[i].cap > 0) { dis[v] = dis[u] + 1; q[++R] = v; if (v == T) return 1; } } } return 0;}int dfs(int u,int flow) { if (u == T) return flow; int used = 0, t; for (int &i = cur[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == dis[u] + 1 && e[i].cap > 0) { t = dfs(v, min(e[i].cap, flow - used)); if (t > 0) { used += t, e[i].cap -= t, e[i ^ 1].cap += t; if (used == flow) break; } } } if (used != flow) dis[u] = -1; return used; }int dinic() { int ans = 0; while (bfs()) ans += dfs(S, INF); return ans;}void solve() { En = 1; memset(head, 0, sizeof(head)); int n = read(), m = read(); S = 0, T = n + n + 1; for (int i = 1; i <= n; ++i) add_edge(i, i + n, 1); for (int i = 1; i <= m; ++i) { int u = read(), v = read(); if (u < 1 || v < 1 || u > n || v > n) continue; // 注意！！！ add_edge(u + n, v, 1); add_edge(v + n, u, 1); } add_edge(S, 2 + n, 2); add_edge(1 + n, T, 1); add_edge(3 + n, T, 1); puts(dinic() == 2 ? "YES" : "NO");}int main() { for (int T = read(); T --; ) solve(); return 0;}
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POJ 1637 Sightseeing tour

题意：混合图欧拉回路的判定。

分析：首先给无向图随便定向，然后判断每个点的初度-入度有没有奇数，如有有，无解。否则S向出>入的点连，容量为(出-入)/2，入>出的点向T连，容量为(入-出)/2，原图中的无向边保留，有向边删掉，跑最大流，有解的条件是满流。

理解：每个的点度数之差如果是奇数，无论怎么改变无向边的方向，这个点的度数之差始终是奇数，所以无解；否则可以尝试着修改一些无向边的方向，使得每个点的出度等于入度。那么在网络中，如果一个点x的出度>入度，那么说明这个点需要改变一些出边变成入边，来满足条件，改变的条数是c=(出-入)/2；如果点y的入度>出度，那么就是改变入边，条数d=(入-出)/2。S向x连一条容量为c的边，y向T连一条容量为d的边。网络流中的一条增广路径经过的边，就是需要反转的边，如果满流，说明存在解。

代码：

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 3005, INF = 1e9;struct Edge{ int to, nxt, cap;}e[N << 1];int head[N], cur[N], dis[N], q[N], En = 1, S, T;void add_edge(int u,int v,int w) { ++En; e[En].to = v; e[En].nxt = head[u]; e[En].cap = w; head[u] = En; ++En; e[En].to = u; e[En].nxt = head[v]; e[En].cap = 0; head[v] = En;}bool bfs() { for (int i = 0; i <= T; ++i) dis[i] = -1, cur[i] = head[i]; int L = 1, R = 0; q[++R] = S; dis[S] = 0; while (L <= R) { int u = q[L ++]; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == -1 && e[i].cap > 0) { dis[v] = dis[u] + 1; q[++R] = v; if (v == T) return 1; } } } return 0;}int dfs(int u,int flow) { if (u == T) return flow; int used = 0, t; for (int &i = cur[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == dis[u] + 1 && e[i].cap > 0) { t = dfs(v, min(e[i].cap, flow - used)); if (t > 0) { used += t, e[i].cap -= t, e[i ^ 1].cap += t; if (used == flow) break; } } } if (used != flow) dis[u] = -1; return used; }int dinic() { int ans = 0; while (bfs()) ans += dfs(S, INF); return ans;}int x[N], y[N], d[N], ru[N], chu[N];void init() { En = 1; memset(head, 0, sizeof(head)); memset(ru, 0, sizeof(ru)); memset(chu, 0, sizeof(chu));}void solve() { init(); int n = read(), m = read(), sum = 0; for (int i = 1; i <= m; ++i) { x[i] = read(), y[i] = read(), d[i] = read(); ru[y[i]] ++, chu[x[i]] ++; } for (int i = 1; i <= n; ++i) { if (abs(ru[i] - chu[i]) & 1) { puts("impossible"); return ; } } S = 0, T = n + 1; for (int i = 1; i <= m; ++i) { if (!d[i]) add_edge(x[i], y[i], 1); } for (int i = 1; i <= n; ++i) { if (chu[i] == ru[i]) continue; if (ru[i] > chu[i]) add_edge(i, T, (ru[i] - chu[i]) / 2); if (chu[i] > ru[i]) add_edge(S, i, (chu[i] - ru[i]) / 2), sum += (chu[i] - ru[i]) / 2; } puts(dinic() == sum ? "possible" : "impossible");}int main() { for (int T = read(); T -- ; ) solve(); return 0;}
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2756: [SCOI2012]奇怪的游戏

分析：黑白染色，讨论，网络流判断。

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 2005;const LL INF = 1e16;struct Edge{ int to, nxt;LL cap;}e[50005];int head[N], cur[N], dis[N], q[N], En = 1, S, T, n, m;void add_edge(int u,int v,LL w) {// printf("%d %d %I64d\n", u, v, w); ++En; e[En].to = v; e[En].nxt = head[u]; e[En].cap = w; head[u] = En; ++En; e[En].to = u; e[En].nxt = head[v]; e[En].cap = 0; head[v] = En;}bool bfs() { for (int i = 0; i <= T; ++i) dis[i] = -1, cur[i] = head[i]; int L = 1, R = 0; q[++R] = S; dis[S] = 0; while (L <= R) { int u = q[L ++]; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == -1 && e[i].cap > 0) { dis[v] = dis[u] + 1; q[++R] = v; if (v == T) return 1; } } } return 0;}LL dfs(int u,LL flow) { if (u == T) return flow; LL used = 0, t; for (int &i = cur[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == dis[u] + 1 && e[i].cap > 0) { t = dfs(v, min(e[i].cap, flow - used)); if (t > 0) { used += t, e[i].cap -= t, e[i ^ 1].cap += t; if (used == flow) break; } } } if (used != flow) dis[u] = -1; return used; }LL dinic() { LL ans = 0; while (bfs()) ans += dfs(S, INF); return ans;}LL a[50][50];int id[50][50];bool check(LL k) { LL t = 0; for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) { if ((i + j) & 1) add_edge(id[i][j], T, k - a[i][j]); else { add_edge(S, id[i][j], k - a[i][j]); t += k - a[i][j]; if (i > 1) add_edge(id[i][j], id[i - 1][j], INF); if (i < n) add_edge(id[i][j], id[i + 1][j], INF); if (j > 1) add_edge(id[i][j], id[i][j - 1], INF); if (j < m) add_edge(id[i][j], id[i][j + 1], INF); } } t = (dinic() == t); En = 1; for (int i = 0; i <= T; ++i) head[i] = 0; return t;}void solve() { n = read(), m = read(); LL Mx = 0; for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) a[i][j] = read(), Mx = max(Mx, a[i][j]); LL sum0 = 0, sum1 = 0, cnt0 = 0, cnt1 = 0; for (int i = 1; i <= n; ++i) { for (int j = 1; j <= m; ++j) { if ((i + j) & 1) sum1 += a[i][j], cnt1 ++; else sum0 += a[i][j], cnt0 ++; } } if (cnt0 == cnt1 && sum0 != sum1) { puts("-1"); return ; } S = 0, T = n * m + 1; for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) id[i][j] = (i - 1) * m + j; if (cnt0 == cnt1) { LL L = Mx, R = INF, ans = 0; while (L <= R) { LL mid = (L + R) >> 1; if (check(mid)) ans = mid, R = mid - 1; else L = mid + 1; } printf("%lld\n", (ans * cnt0 - sum0)); } else { LL k = (sum0 - sum1) / (cnt0 - cnt1); if (k >= Mx && check(k)) printf("%lld\n", (k * cnt0 - sum0)); else puts("-1"); }}int main() { for (int T = read(); T --; ) solve(); return 0;}
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UVA 11082 Matrix Decompressing

分析：经典问题，行列拆成一个点，S->行点，容量为行的和，列点->T，容量为列的和，中间两两连边，注意一下，原题中不能有0，权值范围为[1,20]，因为所有边的容量一样，可以同时减去1（注意S->行，列->T的边也要减），然后求出后统一+1。

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 2005, INF = 1e9;struct Edge{ int to, nxt, cap;}e[N << 1];int head[N], cur[N], dis[N], q[N], En = 1, S, T;void add_edge(int u,int v,int w) { ++En; e[En].to = v; e[En].nxt = head[u]; e[En].cap = w; head[u] = En; ++En; e[En].to = u; e[En].nxt = head[v]; e[En].cap = 0; head[v] = En;}bool bfs() { for (int i = 0; i <= T; ++i) dis[i] = -1, cur[i] = head[i]; int L = 1, R = 0; q[++R] = S; dis[S] = 0; while (L <= R) { int u = q[L ++]; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == -1 && e[i].cap > 0) { dis[v] = dis[u] + 1; q[++R] = v; if (v == T) return 1; } } } return 0;}int dfs(int u,int flow) { if (u == T) return flow; int used = 0, t; for (int &i = cur[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] == dis[u] + 1 && e[i].cap > 0) { t = dfs(v, min(e[i].cap, flow - used)); if (t > 0) { used += t, e[i].cap -= t, e[i ^ 1].cap += t; if (used == flow) break; } } } if (used != flow) dis[u] = -1; return used; }int dinic() { int ans = 0; while (bfs()) ans += dfs(S, INF); return ans;}int a[N], b[N], id[N][N];void solve() { int n = read(), m = read(); S = 0, T = n + m + 1; for (int i = 1; i <= n; ++i) { a[i] = read(); add_edge(S, i, a[i] - a[i - 1] - m); } for (int i = 1; i <= m; ++i) { b[i] = read(); add_edge(i + n, T, b[i] - b[i - 1] - n); } for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) add_edge(i, j + n, 19), id[i][j] = En; dinic(); for (int i = 1; i <= n; ++i, puts("")) for (int j = 1; j <= m; ++j) printf("%d ", 1 + e[id[i][j]].cap); puts(""); En = 1; for (int i = 0; i <= T; ++i) head[i] = 0;}int main() { for (int T = read(), i = 1; i <= T; ++i) { printf("Matrix %d\n", i); solve(); } return 0;}
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1927: [Sdoi2010]星际竞速

分析：首先注意到这是一个DAG，每个点要保证只经过一次，可以从任意点出发。类似一个哈密顿回路（如果把瞬移也加入到图中的话）。考虑一个点是从哪里来的，可以从任意点直接瞬移到这里，可以从连向这个点的点走到这里。

1、那么由于从任意点瞬移到这个点的花费是一定的，所以可以直接从S->i+n（这些点设为i+n）连边，花费为瞬移的时间，容量为1。

2、从连向这个点的位置走过来，花费为路径的权值，由于这是一个类似哈密顿路径的东西，每个点也只能出去一次，所以再建一列点（i），表示从每个点出去，S->i，容量为1（表示这个点只能出去一次），花费为0；对于一条边u->v，连边u->v+n，容量为1，花费为边权。

3、u+n->T连边，容量为1，花费为0。u+n表示已经到过u这个点了，容量为1，表示只能由一个点到这个点（哈密顿路径中，要求出度入度都为1）。

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               #define pa pair
                
                 using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 2005, INF = 1e9;struct Edge{ int fr, to, nxt, cap, cost; } e[50005];int head[N], dis[N], pre[N], En = 1, S, T, q[100005]; bool vis[N];inline void add_edge(int u,int v,int f,int w) { ++En; e[En].fr = u, e[En].to = v, e[En].nxt = head[u], e[En].cap = f, e[En].cost = w; head[u] = En; ++En; e[En].fr = v, e[En].to = u, e[En].nxt = head[v], e[En].cap = 0, e[En].cost = -w; head[v] = En;}bool spfa() { for (int i=1; i<=T; ++i) dis[i] = INF, vis[i] = false, pre[i] = 0; int L = 1, R = 0; q[++R] = S, dis[S] = 0, vis[S] = true, pre[S] = 0; while (L <= R) { int u = q[L ++]; vis[u] = false; for (int i=head[u]; i; i=e[i].nxt) { int v = e[i].to; if (dis[v] > dis[u] + e[i].cost && e[i].cap > 0) { dis[v] = dis[u] + e[i].cost; pre[v] = i; if (!vis[v]) q[++R] = v, vis[v] = true; } } } return dis[T] != INF;}void mcf() { LL Flow = 0, Cost = 0; while (spfa()) { int now = INF; for (int i = T; i != S; i = e[pre[i]].fr) now = min(now, e[pre[i]].cap); for (int i = T; i != S; i = e[pre[i]].fr) e[pre[i]].cap -= now, e[pre[i] ^ 1].cap += now; Cost += 1ll * now * dis[T]; Flow += now; } cout << Cost;}int main() { int n = read(), m = read(); S = 0, T = n + n + 1; for (int i = 1; i <= n; ++i) { int x = read(); add_edge(S, i + n, 1, x); add_edge(S, i, 1, 0); add_edge(i + n, T, 1, 0); } for (int i = 1; i <= m; ++i) { int u = read(), v = read(), w = read(); if (u > v) swap(u, v); add_edge(u, v + n, 1, w); } mcf(); return 0;}
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2324: [ZJOI2011]营救皮卡丘

分析：费用流，建模。首先转化为有向图无环图（DAG）的最小权路径覆盖问题，预处理出g[i][j]，表示i->j只经过编号小于j或者i的点，最短路。然后就是一张n*n的竞赛图了（有向），选小于等于k条路径，覆盖所有点，花费最小。S->0，容量为k，花费为0，S-1~n，容量为1，花费为0；对于u->v，连接u->v+n，容量为1，花费为g[i][j]；连接i+n->T，容量为1，花费为0。

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 505, INF = 1e9;int g[N][N];struct Edge{ int fr, to, nxt, cap, cost; }e[500005];int head[N], dis[N], pre[N], q[100005], En = 1, S, T;bool vis[N];inline void add_edge(int u,int v,int f,int w) { ++En; e[En].fr = u, e[En].to = v, e[En].nxt = head[u], e[En].cap = f, e[En].cost = w; head[u] = En; ++En; e[En].fr = v, e[En].to = u, e[En].nxt = head[v], e[En].cap = 0, e[En].cost = -w; head[v] = En;}bool spfa() { for (int i = 0; i <= T; ++i) dis[i] = INF, vis[i] = false, pre[i] = 0; int L = 1, R = 0; dis[S] = 0, q[++R] = S, vis[S] = true, pre[S] = 0; while (L <= R) { int u = q[L ++]; vis[u] = false; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] > dis[u] + e[i].cost && e[i].cap > 0) { dis[v] = dis[u] + e[i].cost; pre[v] = i; if (!vis[v]) q[++R] = v, vis[v] = true; } } } return dis[T] != INF;}int mcf() { int Flow = 0, Cost = 0; while (spfa()) { int now = INF; for (int i = T; i != S; i = e[pre[i]].fr) now = min(now, e[pre[i]].cap); for (int i = T; i != S; i = e[pre[i]].fr) e[pre[i]].cap -= now, e[pre[i] ^ 1].cap += now; Flow += now; Cost += now * dis[T]; } return Cost;}int main() { int n = read(), m = read(), k = read(); memset(g, 0x3f, sizeof(g)); for (int i = 1; i <= n; ++i) g[i][i] = 0; for (int i = 1; i <= m; ++i) { int u = read(), v = read(), w = read(); g[u][v] = g[v][u] = min(g[u][v], w); } for (int k = 0; k <= n; ++k) for (int i = 0; i <= n; ++i) for (int j = 0; j <= n; ++j) if (k <= i || k <= j) g[i][j] = min(g[i][j], g[i][k] + g[k][j]); S = n + n + 1, T = S + 1; add_edge(S, 0, k, 0); for (int i = 1; i <= n; ++i) add_edge(S, i, 1, 0), add_edge(i + n, T, 1, 0); for (int i = 0; i <= n; ++i) for (int j = i + 1; j <= n; ++j) add_edge(i, j + n, 1, g[i][j]); printf("%d\n", mcf()); return 0;}
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CF 277 E. Binary Tree on Plane

分析：每个点拆成两个点，一个表示作为父节点（i），另一个表示作为子节点（i+n）。然后S向i连边，容量为2（表示只能选两个子节点），花费为0；i+n向T连边容量为1，花费为0；如果i可以作为j的父节点，那么i向j+n连边，容量为1，花费为i与j之间的距离。

如果存在解，那么跑完最小费用最大流后，流量要求大于等于n-1。 

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 500005, INF = 1e9;const double DINF = 1e18;struct Edge{ int fr, to, nxt, cap;double cost; }e[N << 1];int head[N], pre[N], q[N], En = 1, S, T, n; double dis[N];bool vis[N];inline void add_edge(int u,int v,int f,double w) { ++En; e[En].fr = u, e[En].to = v, e[En].nxt = head[u], e[En].cap = f, e[En].cost = w; head[u] = En; ++En; e[En].fr = v, e[En].to = u, e[En].nxt = head[v], e[En].cap = 0, e[En].cost = -w; head[v] = En;}bool spfa() { for (int i = 0; i <= T; ++i) dis[i] = DINF, vis[i] = false, pre[i] = 0; int L = 1, R = 0; dis[S] = 0, q[++R] = S, vis[S] = true, pre[S] = 0; while (L <= R) { int u = q[L ++]; vis[u] = false; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] > dis[u] + e[i].cost && e[i].cap > 0) { dis[v] = dis[u] + e[i].cost; pre[v] = i; if (!vis[v]) q[++R] = v, vis[v] = true; } } } return dis[T] != DINF;}void mcf() { int Flow = 0; double Cost = 0; while (spfa()) { int now = INF; for (int i = T; i != S; i = e[pre[i]].fr) now = min(now, e[pre[i]].cap); for (int i = T; i != S; i = e[pre[i]].fr) e[pre[i]].cap -= now, e[pre[i] ^ 1].cap += now; Flow += now; Cost += 1.0 * now * dis[T]; } if (Flow >= n - 1) printf("%.10lf", Cost); else puts("-1");}struct Node{ int x, y; } A[N];int sqr(int x) { return x * x; }double dist(const Node &A, const Node &B) { return sqrt(1.0 * (sqr(A.x - B.x) + sqr(A.y - B.y)));}int main() { n = read(); for (int i = 1; i <= n; ++i) A[i].x = read(), A[i].y = read(); S = 0, T = n + n + 1; for (int i = 1; i <= n; ++i) add_edge(S, i, 2, 0), add_edge(i + n, T, 1, 0); for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) if (A[j].y > A[i].y) add_edge(j, i + n, 1, dist(A[i], A[j])); mcf(); return 0;}
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2879: [Noi2012]美食节

分析：费用提前计算。考虑第i分菜被第j个厨师做，这是他的第k份，一共做了t份，那么这份菜的花费就是(t-k+1)*T[i][j]。由于t是未知的，这样建图还是不太行，换种方法：考虑这个厨师倒数第k份菜是那个，那么它的花费就是k*T[i][j]。所以可以这样建图：每个厨师拆成sum_p个点，S向每个点连边，容量为1费用为0；每个点向每份菜连边，容量为1费用为T[i][j]；每份菜向T连边，容量为p[i]费用为0。这样的话，边太多了，考虑优化，动态加边，每次增广会找到一条从S->厨师（倒数第i时刻的第j个厨师）->菜->T，因为时刻越靠前花费越多，所以一定会先找大的时刻，所以可以一开始加出所有的倒数第一个时刻的厨师，然后每次增广了一条路，找到这个厨师，把这个厨师的下一个时刻建出。

// luogu-judger-enable-o2#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 500005, INF = 1e9;struct Edge{ int fr, to, nxt, cap, cost; }e[N << 1];int head[N], dis[N], pre[N], q[N], cnt[N], id[N], p[N], t[100][200], En = 1, S, T, n, m, Sum, Index;bool vis[N];inline void add_edge(int u,int v,int f,int w) { ++En; e[En].fr = u, e[En].to = v, e[En].nxt = head[u], e[En].cap = f, e[En].cost = w; head[u] = En; ++En; e[En].fr = v, e[En].to = u, e[En].nxt = head[v], e[En].cap = 0, e[En].cost = -w; head[v] = En;}bool spfa() { for (int i = 0; i <= Index; ++i) dis[i] = INF, vis[i] = false, pre[i] = 0; int L = 1, R = 0; dis[S] = 0, q[++R] = S, vis[S] = true, pre[S] = 0; while (L <= R) { int u = q[L ++]; vis[u] = false; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] > dis[u] + e[i].cost && e[i].cap > 0) { dis[v] = dis[u] + e[i].cost; pre[v] = i; if (!vis[v]) q[++R] = v, vis[v] = true; // vis[v]=true!!! } } } return dis[T] != INF;}void mcf() { int Flow = 0, Cost = 0; while (spfa()) { int now = INF, x; for (int i = T; i != S; i = e[pre[i]].fr) { now = min(now, e[pre[i]].cap); if (e[pre[i]].fr == S) x = e[pre[i]].to; } for (int i = T; i != S; i = e[pre[i]].fr) e[pre[i]].cap -= now, e[pre[i] ^ 1].cap += now; Flow += now; Cost += now * dis[T]; if (Flow == Sum) { cout << Cost; return ; } x = id[x]; cnt[x] ++; if (cnt[x] <= Sum) { ++Index; id[Index] = x; add_edge(S, Index, 1, 0); for (int j = 1; j <= n; ++j) add_edge(Index, j, 1, t[j][x] * cnt[x]); } }}int main() { n = read(), m = read(); for (int i = 1; i <= n; ++i) p[i] = read(), Sum += p[i]; for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) t[i][j] = read(); S = 0, T = m + n + 1; for (int i = 1; i <= n; ++i) add_edge(i, T, p[i], 0); for (int i = 1; i <= m; ++i) add_edge(S, i + n, 1, 0), id[i + n] = i, cnt[i] = 1; for (int i = 1; i <= m; ++i) for (int j = 1; j <= n; ++j) add_edge(i + n, j, 1, t[j][i]); Index = T; mcf(); return 0;}
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HDU 3667 Transportation

分析：每条边差分后拆成5条，分别是a,3a,5a,7a,9a。然后最小费用最大流。

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 105, INF = 1e9;struct Edge{ int fr, to, nxt, cap, cost; }e[60005];int head[N], dis[N], pre[N], q[100005], En = 1, S, T, n, m, k;bool vis[N];inline void add_edge(int u,int v,int f,int w) { ++En; e[En].fr = u, e[En].to = v, e[En].nxt = head[u], e[En].cap = f, e[En].cost = w; head[u] = En; ++En; e[En].fr = v, e[En].to = u, e[En].nxt = head[v], e[En].cap = 0, e[En].cost = -w; head[v] = En;}bool spfa() { for (int i = 0; i <= T; ++i) dis[i] = INF, vis[i] = false, pre[i] = 0; int L = 1, R = 0; dis[S] = 0, q[++R] = S, vis[S] = true, pre[S] = 0; while (L <= R) { int u = q[L ++]; vis[u] = false; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] > dis[u] + e[i].cost && e[i].cap > 0) { dis[v] = dis[u] + e[i].cost; pre[v] = i; if (!vis[v]) q[++R] = v, vis[v] = true; } } } return dis[T] != INF;}void mcf() { int Flow = 0, Cost = 0; while (spfa()) { int now = INF; for (int i = T; i != S; i = e[pre[i]].fr) now = min(now, e[pre[i]].cap); for (int i = T; i != S; i = e[pre[i]].fr) e[pre[i]].cap -= now, e[pre[i] ^ 1].cap += now; Flow += now; Cost += now * dis[T]; } if (Flow != k) puts("-1"); else printf("%d\n", Cost);}void solve(){ S = 0, T = n; add_edge(S, 1, k, 0); for (int i = 1; i <= m; ++i) { int u = read(), v = read(), a = read(), c = read(); for (int j = 1; j <= c; ++j) add_edge(u, v, 1, (j * 2 - 1) * a); } mcf(); En = 1; for (int i = 0; i <= T; ++i) head[i] = 0;}int main() { while (~scanf("%d%d%d", &n, &m, &k)) solve(); return 0;}
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UVA 12092 Paint the Roads 

分析：每个点存在于k个环中，转化为每个点的出度，入度都等于k。然后拆成两列点，S->i，容量为k花费为0；i->j+n，容量为1花费为边的长度；i+n->T，容量为k花费为0。最小费用最大流。

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 105, INF = 1e9; struct Edge{ int fr, to, nxt, cap, cost; }e[10005];int head[N], dis[N], pre[N], q[10005], En = 1, S, T, n, m, k, Sum;bool vis[N];inline void add_edge(int u,int v,int f,int w) { ++En; e[En].fr = u, e[En].to = v, e[En].nxt = head[u], e[En].cap = f, e[En].cost = w; head[u] = En; ++En; e[En].fr = v, e[En].to = u, e[En].nxt = head[v], e[En].cap = 0, e[En].cost = -w; head[v] = En;}bool spfa() { for (int i = 0; i <= T; ++i) dis[i] = INF, vis[i] = false, pre[i] = 0; int L = 1, R = 0; dis[S] = 0, q[++R] = S, vis[S] = true, pre[S] = 0; while (L <= R) { int u = q[L ++]; vis[u] = false; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] > dis[u] + e[i].cost && e[i].cap > 0) { dis[v] = dis[u] + e[i].cost; pre[v] = i; if (!vis[v]) q[++R] = v, vis[v] = true; } } } return dis[T] != INF;}void mcf() { int Flow = 0, Cost = 0; while (spfa()) { int now = INF; for (int i = T; i != S; i = e[pre[i]].fr) now = min(now, e[pre[i]].cap); for (int i = T; i != S; i = e[pre[i]].fr) e[pre[i]].cap -= now, e[pre[i] ^ 1].cap += now; Flow += now; Cost += now * dis[T]; } if (Flow == Sum) printf("%d\n", Cost); else puts("-1");}void solve() { n = read(), m = read(), k = read(); S = 0, T = n + n + 1; Sum = n * k; for (int i = 1; i <= n; ++i) add_edge(S, i, k, 0), add_edge(i + n, T, k, 0); for (int i = 1; i <= m; ++i) { int u = read() + 1, v = read() + 1, d = read(); add_edge(u, v + n, 1, d); } mcf(); En = 1; for (int i = 0; i <= T; ++i) head[i] = 0;}int main() { for (int T = read(); T--; ) solve(); return 0;}
View Code

 

POJ 2175 Evacuation Plan

题意：n个建筑物，每个建筑物里有ai个人，m个避难所，每个避难所可以容纳bi个人，现在给出一个分配方案（g[i][j]表示第i个建筑物去第j个避难所的人数），问是否总的距离是否是最小的，如果不是，输出任意一组比当前优秀的解。

分析：消圈定理：所谓消圈定理，就是在某个流f中，如果其对应的残余网络没有负圈（剩余流量为0的边视为不存在），那它一定就是当前流量下的最小费用流。反之亦然。即「f是最小费用流等价于其残余网络中没有负圈」。by sengxian。那么在这道题中，建出残余网络，找一下负圈，如果存在，那么是说明存在更优的方案，然后在圈上增广一下，得到更优的方案。否则，不存在更优的方案。

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               using namespace std;typedef long long LL;inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;}const int N = 205, INF = 1e9;struct Edge{ int to, nxt, cap, cost; }e[100005];struct Node{ int x, y, z; } A[N], B[N];int g[N][N], cnt[N], head[N], dis[N], pre[N], deg[N], En = 1, S, T, n, m;bool vis[N];queue
                
                  q;inline void add_edge(int u,int v,int f,int w) { ++En; e[En].to = v, e[En].nxt = head[u], e[En].cap = f, e[En].cost = w; head[u] = En;}int spfa() { for (int i = 0; i <= T; ++i) dis[i] = INF, vis[i] = false, deg[i] = pre[i] = 0; q.push(S); dis[S] = 0, vis[S] = true, deg[S] ++; while (!q.empty()) { int u = q.front(); q.pop(), vis[u] = false; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (dis[v] > dis[u] + e[i].cost && e[i].cap > 0) { dis[v] = dis[u] + e[i].cost; pre[v] = u; if (!vis[v]) { q.push(v), vis[v] = true; if ((++deg[v]) > T) return v; } } } } return -1;}void solve(int x) { for (int i = 0; i <= T; ++i) vis[i] = 0; while (!vis[x]) vis[x] = 1, x = pre[x]; // 这里的中间是逗号，写成了分号。。 int z = x, y; do { y = pre[x]; if (y <= n && x > n) g[y][x - n] ++; if (y > n && x <= n) g[x][y - n] --; x = pre[x]; } while (x != z);}int main() { n = read(), m = read(); S = 0, T = n + m + 1; for (int i = 1; i <= n; ++i) A[i].x = read(), A[i].y = read(), A[i].z = read(); for (int i = 1; i <= m; ++i) B[i].x = read(), B[i].y = read(), B[i].z = read(); for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) { g[i][j] = read(); cnt[j] += g[i][j]; int c = abs(A[i].x - B[j].x) + abs(A[i].y - B[j].y) + 1; add_edge(i, j + n, INF - g[i][j], c); add_edge(j + n, i, g[i][j], -c); } for (int i = 1; i <= n; ++i) { add_edge(S, i, A[i].z, 0); add_edge(i, S, 0, 0); } for (int i = 1; i <= m; ++i) { add_edge(i + n, T, B[i].z - cnt[i], 0); add_edge(T, i + n, cnt[i], 0); } int st = spfa(); if (st == -1) { puts("OPTIMAL"); return 0; } solve(st); puts("SUBOPTIMAL"); for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) printf("%d%c", g[i][j], j == m ? '\n' : ' '); return 0;}
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备注：缺一道JOJ 2453，没找到提交地址？题目挺好的。

转载于:https://www.cnblogs.com/mjtcn/p/10274547.html

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