關于有源上下界最大流: https://blog.csdn.net/regina8023/article/details/45815023

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n, m, k, l, r, s, t, superS, superT; const int MAXN = 5000;//點數的最大值 const int MAXM = 120000;//邊數的最大值 const int INF = 0x3f3f3f3f; struct Edge {int to,next,cap,flow;} edge[MAXM]; //注意是MAXM int tol; int head[MAXN]; void init() {tol = 2;memset(head,-1,sizeof(head)); } void addedge(int u,int v,int w,int rw = 0) {edge[tol].to = v;edge[tol].cap = w;edge[tol].flow = 0;edge[tol].next = head[u];head[u] = tol++;edge[tol].to = u;edge[tol].cap = rw;edge[tol].flow = 0;edge[tol].next = head[v];head[v] = tol++; } int Q[MAXN]; int dep[MAXN],cur[MAXN],sta[MAXN]; bool bfs(int s,int t,int n) {int front = 0,tail = 0;memset(dep,-1,sizeof(dep[0])*(n+1));dep[s] = 0;Q[tail++] = s;while(front < tail){int u = Q[front++];for(int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].to;if(edge[i].cap > edge[i].flow && dep[v] == -1){dep[v] = dep[u] + 1;if(v == t)return true;Q[tail++] = v;}}}return false; } int dinic(int s,int t,int n) {int maxflow = 0;while(bfs(s,t,n)){for(int i = 0; i < n; i++)cur[i] = head[i];int u = s, tail = 0;while(cur[s] != -1){if(u == t){int tp = INF;for(int i = tail-1; i >= 0; i--)tp = min(tp,edge[sta[i]].cap-edge[sta[i]].flow);maxflow += tp;for(int i = tail-1; i >= 0; i--){edge[sta[i]].flow += tp;edge[sta[i]^1].flow -= tp;if(edge[sta[i]].cap-edge[sta[i]].flow == 0)tail = i;}u = edge[sta[tail]^1].to;}else if(cur[u] != -1 && edge[cur[u]].cap > edge[cur[u]].flow && dep[u] + 1 == dep[edge[cur[u]].to]){sta[tail++] = cur[u];u = edge[cur[u]].to;}else{while(u != s && cur[u] == -1)u = edge[sta[--tail]^1].to;cur[u] = edge[cur[u]].next;}}}return maxflow; } int main() {int kase = 1;while(~scanf("%d %d %d", &m, &n, &k)){scanf("%d %d", &l, &r);s = n + m + 1;t = n + m + 2;superS = n + m + 3;superT = n + m + 4;init();for(int i = 1; i <= k; i++){int u, v;scanf("%d %d", &u , &v);addedge(u, v + n, 1); //二分圖 建一條容量為1的邊 }addedge(t,s,INF); //從匯點向源點建一條inf的邊for(int i = 1; i <= n; i++){addedge(s, i, r - l); //從源點向左半圖連邊 addedge(superS, i, l);addedge(s, superT, l);}for(int i = 1; i <= m; i++){int aim = i + n;addedge(aim, t, r - l);addedge(aim, superT, l);addedge(superS, t, l);}printf("Case %d: ", kase++);int ans = dinic(superS, superT, n+m+4);if(ans == (n+m)*l) puts("Yes");else puts("No");}return 0; }

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轉載于:https://www.cnblogs.com/Jadon97/p/9692975.html

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