【bzoj2039】[2009国家集训队]employ人员雇佣 最小割

刚开始读错题了，把题目想简单了，看了一下Popoqqq大爷的题解。

最小割
假设全部雇佣
i向汇点T连一条容量为ai的边
对于利润(i,j)
源点S向i和j分别连一条容量为eij的边
i和j之间连一条容量为2eij的边
答案为e的和-ans
如果雇佣两个，则不用割掉
如果只雇佣一个，则要割掉2eij
如果两个都不雇佣，则割掉2eij

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#define maxn 1010
#define maxm 4000100
#define inf 1000000000

using namespace std;

int head[maxn],to[maxm],c[maxm],next[maxm],q[maxn],d[maxn];
int e[maxn][maxn];
int n,m,num,s,t,ans;

void addedge(int x,int y,int z)
{
	num++;to[num]=y;c[num]=z;next[num]=head[x];head[x]=num;
	num++;to[num]=x;c[num]=0;next[num]=head[y];head[y]=num;
}

bool bfs()
{
	memset(d,-1,sizeof(d));
	int l=0,r=1;
	q[1]=s;d[s]=0;
	while (l<r)
	{
		int x=q[++l];
		for (int p=head[x];p;p=next[p])
		  if (c[p] && d[to[p]]==-1)
		  {
		  	d[to[p]]=d[x]+1;
		  	q[++r]=to[p];
		  }
	}
	if (d[t]==-1) return 0; else return 1;
}

int find(int x,int low)
{
	if (x==t || low==0) return low;
	int totflow=0;
	for (int p=head[x];p;p=next[p])
	  if (c[p] && d[to[p]]==d[x]+1)
	  {
	  	int a=find(to[p],min(low,c[p]));
	  	c[p]-=a;c[p^1]+=a;
	  	low-=a;totflow+=a;
	  	if (low==0) return totflow;
	  }
	if (low) d[x]=-1;
	return totflow;
}

int main()
{
	scanf("%d",&n);
	num=1;s=0;t=n+1;
	for (int i=1;i<=n;i++)
	{
		int x;
		scanf("%d",&x);
		addedge(i,t,x);
	}
	for (int i=1;i<=n;i++)
	  for (int j=1;j<=n;j++)
	  {
	    scanf("%d",&e[i][j]);
	    ans+=e[i][j];
	  }
	for (int i=1;i<=n;i++)
	  for (int j=i+1;j<=n;j++)
	    addedge(s,i,e[i][j]),addedge(s,j,e[i][j]),addedge(i,j,2*e[i][j]),addedge(j,i,2*e[i][j]);
	while (bfs()) ans-=find(s,inf);
	printf("%d\n",ans);
	return 0;
}