POJ2135 Farm Tour

POJ2135 Farm Tour

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6299		Accepted: 2248

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

**********************************************************************************

题目大意：一个人要从1走到n又要从n走到1，每条路上都有一个花费，要求，同一条路只走一遍以及花费最小。

解题思路：最小费用最大流。因为要从1到n再从n到1，其实可以想成是从1到n走了两次，而这两次里没有一条路是重复的。将路拆成两条，然后每条的容量都是1，费用都是v，求最小费用最大流即可。

PS：万恶的期中考试，去备战了3天，结果现在写代码竟然如此生疏，不太爽。

#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#define N 1005
#define M 40005
#define INF 0x3f3f3f3f
using namespace std;

int head[N],ed[M],val[M],nxt[M],up[M],eid;
int n,m;
int vis[N],dis[N],pre[N],road[N];

void add_edge(int s,int e,int u,int v)
{
    up[eid]=u;
    ed[eid]=e;          val[eid]=v;
    nxt[eid]=head[s];   head[s]=eid++;
    up[eid]=0;
    ed[eid]=s;          val[eid]=-v;
    nxt[eid]=head[e];   head[e]=eid++;
}

void spfa(void)
{
    queue<int>que;
    memset(vis,0,sizeof(vis));
    memset(pre,0,sizeof(pre));
    for(int i=2;i<=n;i++)
        dis[i]=INF;
    dis[1]=0;
    que.push(1);
    while(!que.empty())
    {
        int t=que.front();
        que.pop();
        vis[t]=0;
        for(int i=head[t];~i;i=nxt[i])
        {
            if(up[i]==0)continue;
            int e=ed[i],v=val[i];
            if(dis[t]+v<dis[e])
            {
                dis[e]=dis[t]+v;
                pre[e]=t;
                road[e]=i;
                if(vis[e]==0)
                {
                    vis[e]=1;
                    que.push(e);
                }
            }
        }
    }
}

void re(void)
{
    scanf("%d%d",&n,&m);
    memset(head,-1,sizeof(head));
    eid=0;
    for(int i=1;i<=m;i++)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        add_edge(a,b,1,c);
        add_edge(b,a,1,c);
    }
}

void run(void)
{
    int ans=0;
    for(int h=1;h<=2;h++)
    {
        spfa();
        for(int i=n;i!=1;i=pre[i])
        {
            up[road[i]]-=1;
            up[road[i]^1]+=1;
        }
        ans+=dis[n];
    }
    printf("%d\n",ans);
}

int main()
{
    re();
    run();
}