# ACM/ICPC 之 伞兵-最小割转最大流(POJ3308)

```//以行列建点，伞兵位置为单向边-利用对数将乘积转加法
//最小割转最大流
//Time:63Ms Memory:792K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;

#define MAXN 105
#define INF 100000
#define EPS 1e-7

int n,m,p;
int s,t;
double res[MAXN][MAXN];
int pre[MAXN];

bool bfs()
{
memset(pre,-1,sizeof(pre));
queue<int> q;
q.push(s);  pre[s] = 0;
while(!q.empty()){
int cur = q.front();
q.pop();
for(int i = 1; i <= t; i++)
{
if(pre[i] == -1 && res[cur][i] > EPS)
{
pre[i] = cur;
if(i == t)  return true;
q.push(i);
}
}
}
return false;
}

double EK()
{
double maxFlow = 0;
while(bfs()){
double mind = INF;
for(int i = t; i != s; i = pre[i])
mind = min(mind, res[pre[i]][i]);
for(int i = t; i != s; i = pre[i])
{
res[pre[i]][i] -= mind;
res[i][pre[i]] += mind;
}
maxFlow += mind;
}
return maxFlow;
}

int main()
{
//freopen("in.txt", "r", stdin);

int T;
scanf("%d",&T);
while(T--){
memset(res,0,sizeof(res));
scanf("%d%d%d", &n,&m,&p);
s = 0;  t = n + m + 1;
double c;
for(int i = 1; i <= n; i++)
{
scanf("%lf", &c);
res[s][i] = log(c); //乘法转为加法
}
for(int i = 1; i <= m; i++)
{
scanf("%lf", &c);
res[i+n][t] = log(c);
}
for(int i = 1; i <= p; i++)
{
int rr,cc;
scanf("%d%d", &rr,&cc);
res[rr][cc+n] = INF;
}

printf("%.4f\n", exp(EK()));    //恢复

}

return 0;
}
```

原文作者：Inkblots
原文地址: https://www.cnblogs.com/Inkblots/p/5717162.html
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