poj 2135

一道基础题，却因为这样那样的粗心，用了这么长时间，和uva10086一模一样

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int maxn=1100;
const int maxe=30000;
const int inf=200000000;
struct Edge
{
	int from,to,flow,cap,cost,next;
};
int n,m,tot;
int head[maxn],d[maxn],a[maxn],vis[maxn];
Edge edges[maxe];
void addedge(int from,int to,int cap,int cost)
{
	edges[tot].from=from;edges[tot].to=to;edges[tot].flow=0;edges[tot].cap=cap;edges[tot].cost=cost;
	edges[tot].next=head[from];head[from]=tot;
	tot++;
	edges[tot].from=to;edges[tot].to=from;edges[tot].flow=0;edges[tot].cap=cap;edges[tot].cost=cost;
	edges[tot].next=head[to];head[to]=tot;
	tot++;
}
int mfmc()
{
	int cost=0,flow=0,i;
	int p[maxn];
	queue<int> q;
	for(;;)
	{
		for(i=1;i<=n;i++)
		{
			vis[i]=0;
			d[i]=inf;
		}
		vis[1]=1;d[1]=0;a[1]=inf;
		q.push(1);
		while(!q.empty())
		{
			int u=q.front();q.pop();
			vis[u]=0;
			for(int i=head[u];i!=-1;i=edges[i].next)
			{
				Edge& e=edges[i];
				if(e.cap>e.flow&&d[e.to]>(d[u]+e.cost))
				{
					d[e.to]=d[u]+e.cost;
					p[e.to]=i;
					a[e.to]=min(a[u],e.cap-e.flow);
					if(!vis[e.to])
					{
						vis[e.to]=1;
						q.push(e.to);
					}
				}
			}
		}
		if(d[n]==inf) break;
		flow+=a[n];
		cost+=a[n]*d[n];
		if(flow==2) return cost;
		for(int x=n;x!=1;)
		{
			edges[p[x]].flow+=a[n];
			edges[p[x]^1].flow-=a[n];
			edges[p[x]^1].cost=-edges[p[x]^1].cost;
			x=edges[p[x]].from;
		}
	}
	return -1;
}
int main()
{
	while(cin>>n>>m)
	{
		memset(head,-1,sizeof(head));
		tot=0;
		int i,j,u,v,c;
		for(i=1;i<=m;i++)
		{
			scanf("%d%d%d",&u,&v,&c);
			addedge(u,v,1,c);
		}
		cout<<mfmc()<<endl;
	}
	return 0;
}