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单源点汇点无向图,要阻隔某个点的流量,必须在一个点上消耗一定的价值,问你能否在消耗价值不超过k的前提下,阻隔源点到汇点的流量。
直接对于有权值的点拆点,拆后边容量即为点权。其余的点的容量无穷,最大流即可。
召唤代码君:
#include <iostream> #include <cstring> #include <cstdio> #define maxn 5555 #define maxm 555555 using namespace std; const int inf=11111; int to[maxm],c[maxm],next[maxm],first[maxn],edge; int a[maxn],d[maxn],tag[maxn],TAG=222; bool can[maxn]; int Q[maxm],bot,top; int n,m,l,s,t; void _init() { edge=-1; for (int i=1; i<=n+n; i++) first[i]=-1; } void addedge(int U,int V,int W) { edge++; to[edge]=V,c[edge]=W,next[edge]=first[U],first[U]=edge; edge++; to[edge]=U,c[edge]=0,next[edge]=first[V],first[V]=edge; } bool bfs() { Q[bot=top=1]=t,tag[t]=++TAG,d[t]=0,can[t]=false; while (bot<=top) { int cur=Q[bot++]; for (int i=first[cur]; i!=-1; i=next[i]) if (c[i^1] && tag[to[i]]!=TAG) { tag[to[i]]=TAG,d[to[i]]=d[cur]+1; can[to[i]]=false,Q[++top]=to[i]; if (to[i]==s) return true; } } return false; } int dfs(int cur,int num) { if (cur==t) return num; int tmp=num,k; for (int i=first[cur]; i!=-1; i=next[i]) if (c[i] && d[to[i]]==d[cur]-1 && tag[to[i]]==TAG && !can[to[i]]) { k=dfs(to[i],min(c[i],num)); if (k) num-=k,c[i]-=k,c[i^1]+=k; if (!num) break; } if (num) can[cur]=true; return tmp-num; } int main() { int U,V,Flow,K; while (scanf("%d",&K)!=EOF) { scanf("%d%d%d%d",&n,&m,&s,&t); for (int i=1; i<=n; i++) scanf("%d",&a[i]); _init(); for (int i=1; i<=m; i++) { scanf("%d%d",&U,&V); addedge(U+n,V,inf),addedge(V+n,U,inf); } if (s==t) { puts("NO"); continue; } s+=n; for (int i=1; i<=n; i++) addedge(i,i+n,a[i]); for (Flow=0; Flow<=K && bfs(); bfs()) Flow+=dfs(s,inf); if (K>=Flow) puts("YES"); else puts("NO"); } return 0; }
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原文地址:http://www.cnblogs.com/Canon-CSU/p/3870718.html