POJ 2135 最小费用最大流

思路：
源->1连费用0 流量2
其它的边 费用w 流量1
n->汇 费用0 流量2
最小费用流 搞定~

//By SiriusRen
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 2010
#define M 201000
int n,m,first
,next[M],v[M],edge[M],cost[M],tot,xx,yy,zz,S,T;
int vis
,with
,d
,minn
,ans;
void Add(int x,int y,int C,int E){
edge[tot]=E,cost[tot]=C,v[tot]=y,next[tot]=first[x],first[x]=tot++;
}
void add(int x,int y,int C,int E){Add(x,y,C,E),Add(y,x,-C,0);}
bool tell(){
memset(vis,0,sizeof(vis)),memset(d,0x3f,sizeof(d));
memset(with,0,sizeof(with)),memset(minn,0x3f,sizeof(minn));
queue<int>q;d[S]=0;q.push(S);
while(!q.empty()){
int t=q.front();q.pop();vis[t]=0;
for(int i=first[t];~i;i=next[i])
if(d[v[i]]>d[t]+cost[i]&&edge[i]>0){
d[v[i]]=d[t]+cost[i],minn[v[i]]=min(minn[t],edge[i]),with[v[i]]=i;
if(!vis[v[i]])vis[v[i]]=1,q.push(v[i]);
}
}return d[T]!=0x3f3f3f3f;
}
int zeng(){
for(int i=T;i!=S;i=v[with[i]^1])
edge[with[i]]-=minn[T],edge[with[i]^1]+=minn[T];
return minn[T]*d[T];
}
int main(){
memset(first,-1,sizeof(first));
scanf("%d%d",&n,&m);
S=n+1,T=n+2;
for(int i=1;i<=m;i++){
scanf("%d%d%d",&xx,&yy,&zz);
add(xx,yy,zz,1),add(yy,xx,zz,1);
}add(S,1,0,2),add(n,T,0,2);
while(tell())ans+=zeng();
printf("%d\n",ans);
}