POJ 2135 Farm Tour (最小费用最大流)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<queue>
#include<algorithm>
#define MAX 1000
#define inf 0x3f3f3f3f
using namespace std;

struct node{
int to,next,w,c;
}q[30024*5];

int star,en,head[30024*5],cnt,flow,cost,dis[3024],cur[10024];
bool vis[1010*5];
int f[10024];
void bu(int a,int b,int w,int c){
q[cnt].to=b;
q[cnt].w=w;
q[cnt].c=c;
q[cnt].next=head[a];
head[a]=cnt++;

q[cnt].to=a;
q[cnt].w=0;
q[cnt].c=-c;
q[cnt].next=head[b];
head[b]=cnt++;
}

bool SPFA(){
memset(vis,false,sizeof(vis) );
memset(dis,inf,sizeof(dis) );
dis[star]=0;
cur[star]=-1;
f[star]=inf;
queue<int>qu;
qu.push(star);
while(!qu.empty()){
int u=qu.front();
qu.pop();
vis[u]=false;
for(int i=head[u];~i;i=q[i].next){
int v=q[i].to;
if(dis[v]>dis[u]+q[i].c&&q[i].w){
dis[v]=dis[u]+q[i].c;
f[v]=min(f[u],q[i].w);
cur[v]=i;
if(!vis[v]){
vis[v]=true;
qu.push(v);
}
}
}
}
if(dis[en]==inf)return false;
flow+=f[en];
cost+=f[en]*dis[en];
//printf("%d\n",flow);
for(int i=cur[en];~i;i=cur[q[i^1].to]){
q[i].w-=f[en];
q[i^1].w+=f[en];
}
return true;
}

int main(){
int n,m,i,j,k,a,b,c;
while(~scanf("%d%d",&n,&m)){
cnt =0;
memset(head,-1,sizeof(head));
for(i = 0;i <m;++ i){
scanf("%d%d%d",&a,&b,&c);
bu(a,b,1,c);
bu(b,a,1,c);
}
star = 0,en = n+1;
cost = flow = 0;
bu(star,1,2,0);
bu(n,n+1,2,0);
while(SPFA());
printf("%d\n",cost);
}
return 0;
}