hdu 3061 最大流最小割

题目:http://acm.hdu.edu.cn/showproblem.php?pid=3061 中文题

最小割，增加源和汇，把源与权为正的点连上界为该点权值的边，权为负的点与汇连上上界为该点权值相反数的边，然后如果攻占a必须先占b，则连上权值无穷的边a->b，所有正值之和减去最大流，即答案。。。

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<memory.h>
using namespace std;
const int M=502;
const int MAXN=500000;
const int INF=0x3f3ff3f;
int t,n,m,tot;
int gap[M],dis[M],pre[M],head[M],cur[M];
int NE,NV,sink,s[M],e[M],tol[M*2];
struct Node
{
int c,pos,next;
} E[MAXN];
#define FF(i,NV) for(int i=0;i<NV;i++)
int sap(int s,int t)
{
memset(dis,0,sizeof(int)*(NV+1));
memset(gap,0,sizeof(int)*(NV+1));
FF(i,NV) cur[i] = head[i];
int u = pre[s] = s,maxflow = 0,aug =INF;
gap[0] = NV;
while(dis[s] < NV)
{
loop:
for(int &i = cur[u]; i != -1; i = E[i].next)
{
int v = E[i].pos;
if(E[i].c && dis[u] == dis[v] + 1)
{
aug=min(aug,E[i].c);
pre[v] = u;
u = v;
if(v == t)
{
maxflow += aug;
for(u = pre[u]; v != s; v = u,u = pre[u])
{
E[cur[u]].c -= aug;
E[cur[u]^1].c += aug;
}
aug =INF;
}
goto loop;
}
}
if( (--gap[dis[u]]) == 0)   break;
int mindis = NV;
for(int i = head[u]; i != -1 ; i = E[i].next)
{
int v = E[i].pos;
if(E[i].c && mindis > dis[v])
{
cur[u] = i;
mindis = dis[v];
}
}
gap[ dis[u] = mindis+1 ] ++;
u = pre[u];
}
return maxflow;
}
void addEdge(int u,int v,int c )
{
E[NE].c = c;
E[NE].pos = v;
E[NE].next = head[u];
head[u] = NE++;
E[NE].c = 0;
E[NE].pos = u;
E[NE].next = head[v];
head[v] = NE++;
}
int main()
{
int n, m;
int  sum, source, sink, vn,a,b;
while(scanf("%d %d", &n, &m)!=EOF)
{
NE=0,tot=0;
sum = 0;
source = 0;
sink=n+1;
NV=sink+1;
memset(head, -1, sizeof(head));
for(int i=1;i<=n;i++)
{
scanf("%d",&vn);
if(vn>0)
{
addEdge(i,sink,vn);
sum+=vn;
}
else addEdge(source,i,-vn);
}
for(int i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
addEdge(b,a,INF);
}
printf("%d\n",sum-sap(source,sink));
}
return 0;
}