poj2135 Farm Tour

Farm Tour

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18334		Accepted: 7095

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

Source

USACO 2003 February Green

大致题意：从1到n求两条不相交的路径，使得边权和最大.

分析：第一次做有点难想到.如果不要求不相交，那么直接最短路就好了.要求不相交，每条边最多只能走一次，那么可以给每个边加一个容量1，最后求起点到终点流量为2的最小费用流.套模板.思路比较巧妙。

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 500010,inf = 0x3f3f3f3f;
int n,m,head[maxn],to[maxn],nextt[maxn],tot = 2,pre[maxn],w[maxn],c[maxn],ans;
int d[maxn],vis[maxn],S,T,vis2[maxn];

void add(int x,int y,int z,int p)
{
    w[tot] = z;
    c[tot] = p;
    to[tot] = y;
    nextt[tot] = head[x];
    head[x] = tot++;
}

bool spfa()
{
    memset(d,inf,sizeof(d));
    memset(vis2,0,sizeof(vis2));
    d[S] = 0;
    queue <int> q;
    q.push(S);
    memset(vis,0,sizeof(vis));
    vis[S] = 1;
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u];i;i = nextt[i])
        {
            int v = to[i];
            if (w[i] && d[v] > d[u] + c[i])
            {
                d[v] = d[u] + c[i];
                if (!vis[v])
                {
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return d[T] < inf;
}

int dfs(int u,int f)
{
    if (u == T)
    {
        ans += f * d[T];
        return f;
    }
    int res = 0;
    vis2[u] = 1;
    for (int i = head[u];i;i = nextt[i])
    {
        int v = to[i];
        if (!vis2[v] && w[i] && d[v] == d[u] + c[i])
        {
            int temp = dfs(v,min(f - res,w[i]));
            w[i] -= temp;
            w[i ^ 1] += temp;
            res += temp;
            if (res == f)
                return f;
        }
    }
    return res;
}

void dinic()
{
    while(spfa())
        dfs(S,inf);
}

int main()
{
    scanf("%d%d",&n,&m);
    S = 0,T = n + 1;
    for (int i = 1; i <= m; i++)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        add(a,b,1,c);
        add(b,a,0,-c);
        add(b,a,1,c);
        add(a,b,0,-c);
    }
        add(S,1,2,0);
        add(1,S,0,0);
        add(n,T,2,0);
        add(T,n,0,0);
    dinic();
    printf("%d\n",ans);

    return 0;
}