# POJ 2135 Farm Tour （最小费用流）

Farm Tour

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 15574 Accepted: 6002

Description

When FJ’s friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000.To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again.

He wants his tour to be as short as possible, however he doesn’t want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm.

Input

* Line 1: Two space-separated integers: N and M.* Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path’s length.

Output

A single line containing the length of the shortest tour.

Sample Input

4 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2

Sample Output

6

### 题意

FJ有N个农场，M条路，FJ要领朋友游玩，从1走到N，再回到1，不走重复路，每条路长度不一样，问最短路长为多少。

### AC代码

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<vector>
#include<queue>
using namespace std;

const int MAXX = 40010;
const int INF = 0x3f3f3f3f;

struct edge
{
int to;
int next;
int cap;
int flow;
int cost;
} edge[MAXX];

int pre[MAXX],dis[MAXX];
bool vis[MAXX];
int N;  //节点总个数

void init(int n)
{
N=n;
tol=0;
}

void addedge(int u,int v,int cap,int cost)
{
edge[tol].to=v;
edge[tol].cap=cap;
edge[tol].cost=cost;
edge[tol].flow=0;
edge[tol].to=u;
edge[tol].cap=0;
edge[tol].cost=-cost;
edge[tol].flow=0;
}

bool spfa(int s,int t)
{
queue<int>q;
for(int i=0; i<N; i++)
{
dis[i]=INF;
vis[i]=false;
pre[i]=-1;
}
dis[s]=0;
vis[s]=true;
q.push(s);
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u]=false;
{
int v=edge[i].to;
if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost)
{
dis[v]=dis[u]+edge[i].cost;
pre[v]=i;
if(!vis[v])
{
vis[v]=true;
q.push(v);
}
}
}
}
if(pre[t]==-1)return false;
else return true;
}

/*
* int s 起点
* int t 终点
* 返回最大流
*/
int minCostMaxFlow(int s,int t)
{
int cost=0;
while(spfa(s,t))
{
int minn=INF;
for(int i=pre[t]; i!=-1; i=pre[edge[i^1].to])
{
if(minn>edge[i].cap-edge[i].flow)
minn=edge[i].cap-edge[i].flow;
}
for(int i=pre[t]; i!=-1; i=pre[edge[i^1].to])
{
edge[i].flow+=minn;
edge[i^1].flow-=minn;
cost+=edge[i].cost*minn;
}
}
return cost;
}

int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
init(n+2);  //因为有增加的两个点
int st,ed,cost;
for(int i=0; i<m; i++)
{
scanf("%d%d%d",&st,&ed,&cost);