HDU 4289 最小割=最大流 求去掉最少点权值使得 起末点不连通

题意：

n个点 m条边

下面起点 和终点

n行表示点权值

m条无向边

 

问：

去掉一些点需要的花费为该点的点权值，问要最少多少花费可以使得起点 和 终点 不连通

 

网络流裸题，按题目直接可以建图；

 

#include <stdio.h>#include <algorithm>#include <queue>#include <vector>using namespace std;#define ll int#define N 80000#define M 401#define inf 536870912inline int Max(int a,int b){return a<b?b:a;}inline int Min(int a,int b){return a>b?b:a;}struct Edge{int from, to, cap, nex;}edge[N];int head[M], edgenum;void addedge(int u, int v, int cap){Edge E = { u, v, cap, head[u]};edge[ edgenum ] = E;head[u] = edgenum ++;Edge E1= { v, u, 0,   head[v]};edge[ edgenum ] = E1;head[v] = edgenum ++;}int sign[M], s, t;bool BFS(int from, int to){memset(sign, -1, sizeof(sign));sign[from] = 0;queue<int>q;q.push(from);while( !q.empty() ){int u = q.front(); q.pop();for(int i = head[u]; i!=-1; i = edge[i].nex){int v = edge[i].to;if(sign[v]==-1 && edge[i].cap){sign[v] = sign[u] + 1, q.push(v);if(sign[to] != -1)return true;}}}return false;}int Stack[N], top, cur[N];int dinic(){int ans = 0;while( BFS(s, t) ){memcpy(cur, head, sizeof(head));int u = s;top = 0;while(1){if(u == t){int flow = inf, loc;//loc 表示 Stack 中 cap 最小的边for(int i = 0; i < top; i++)if(flow > edge[ Stack[i] ].cap){flow = edge[Stack[i]].cap;loc = i;}for(int i = 0; i < top; i++){edge[ Stack[i] ].cap -= flow;edge[Stack[i]^1].cap += flow;}ans += flow;top = loc;u = edge[Stack[top]].from;}for(int i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;if(cur[u] != -1){Stack[top++] = cur[u];u = edge[ cur[u] ].to;}else{if( top == 0 )break;sign[u] = -1;u = edge[ Stack[--top] ].from;}}}return ans;}int n;int main(){ll i,u,v,temp,m;while(~scanf("%d %d",&n,&m)){memset(head, -1, sizeof(head)); edgenum = 0;scanf("%d %d",&s,&t);s+=n;for(i=1;i<=n;i++){scanf("%d",&temp);addedge(i+n,i, temp);}while(m--){scanf("%d %d",&u,&v);addedge(u, v+n, inf);addedge(v, u+n, inf);}printf("%d\n", dinic());}return 0;}/*5 65 35234121 55 42 32 44 32 1*/