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【题目链接】
http://acm.hdu.edu.cn/showproblem.php?pid=4289
题目意思
有n个城市,m条边。每个城市用有花费权值。花费对应权值可以去除这个城市问断绝s-t连通的最少花费
解题思路
最小割值上等于最大流,把每个城市拆成p,p’.<p,p’>权值为城市权值,其他边p进p’出权值为无限大
代码部分
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <string>
#include <map>
using namespace std;
#define inf 0x3f3f3f3
#define LL long long
#define pii pair<int ,int>
const int N = 450;
int dis[N];
int mp[N][N];
int n,m,s,t;
bool bfs ()
{
memset(dis,-1,sizeof(dis));
dis[s] = 0;
queue<int> q;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
for (int i = 1; i <= 2*n; i++)
{
if (dis[i] == -1 && mp[u][i] > 0)
{
dis[i] = dis[u]+1;
q.push(i);
}
}
}
return dis[t+n] != -1;
}
int dfs(int v,int flow)
{
if (v == t+n)
return flow;
int delta = flow;
for (int i = 1; i <= 2*n; i++)
{
if(dis[i]==dis[v]+1 && mp[v][i] > 0)
{
int a = dfs(i,min(delta,mp[v][i]));
if(a)
{
mp[v][i] -= a;
mp[i][v] += a;
delta -= a;
if (delta == 0)
return flow;
}
}
}
return flow-delta;
}
int dinic()
{
int ans = 0;
while(bfs())
{
ans += dfs(s,inf);
}
return ans;
}
int main()
{
int v,u,w;
while(~scanf("%d %d",&n,&m))
{
memset(mp,0,sizeof(mp));
scanf("%d %d",&s,&t);
for (int i = 1; i <= n; i++)
{
scanf("%d",&w);
mp[i][i+n] = w;
}
for (int i = 1; i <= m; i++)
{
scanf("%d %d",&u,&v);
mp[u+n][v] = inf;
mp[v+n][u] = inf;
}
cout<<dinic()<<endl;
}
return 0;
}