Control HDU

题目链接: Control HDU - 4289

题目大意

一张无向图, n个顶点, m条边, 现在要去除某些顶点, 使得节点s和节点d之间没有任何路径相连, 每个顶点有一个cost, 去除这个顶点的花费, 求最小花费

思路

将每个顶点v拆成两个顶点v1,v2, 之间连一条容量为去除这个顶点的cost, 求s1,d2之间的最小割

代码

#include <bits/stdc++.h>using namespace std;const int MAXV = 500, INF = 0X3F3F3F3F;struct edge{    int to, cap, rev;    edge(int a, int b, int c) :to(a), cap(b), rev(c){}};vector<edge> G[MAXV];int iter[MAXV], level[MAXV], S, T;void init(){    for(int i=0; i<MAXV; ++i) G[i].clear();}void add_edge(int from, int to, int cap){    G[from].push_back(edge(to, cap, G[to].size()));    G[to].push_back(edge(from, 0, G[from].size()-1));}bool bfs(){    memset(level, -1, sizeof(level));    level[S] = 0;    queue<int> que;    que.push(S);    while(!que.empty())    {        int v = que.front(); que.pop();        for(int i=0; i<(int)G[v].size(); ++i)        {            edge &e = G[v][i];            if(e.cap>0 && level[e.to]==-1)            {                level[e.to] = level[v] + 1;                que.push(e.to);            }        }    }    return level[T]!=-1;}int dfs(int v, int f){    if(v == T) return f;    for(int &i=iter[v]; i<(int)G[v].size(); ++i)    {        edge &e = G[v][i];        if(e.cap>0 && level[e.to]>level[v])        {            int d = dfs(e.to, min(f, e.cap));            if(d)            {                e.cap -= d;                G[e.to][e.rev].cap+= d;                return d;            }        }    }    return 0;}int max_flow(){    int flow = 0;    while(bfs())    {        memset(iter, 0, sizeof(iter));        int f;        while((f=dfs(S, INF))) flow += f;    }    return flow;}int n, m, s, d;int main(){    while(scanf("%d%d", &n, &m) == 2)    {        init();        scanf("%d%d", &s, &d);        S = s-1;        T = n+d-1;        int c;        for(int i=0; i<n; ++i)        {            scanf("%d", &c);            add_edge(i, i+n, c);        }        int a, b;        for(int i=0; i<m; ++i)        {            scanf("%d%d", &a, &b);            a--, b--;            add_edge(a+n, b, INF);            add_edge(b+n, a, INF);        }        printf("%d\n", max_flow());    }    return 0;}