【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 1000000000using 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;}