Farm Tour(最小费用最大流模板)

Farm Tour

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18150		Accepted: 7023

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

 

 

//题意：n 个点， m 条无向边，要求从 1 走到 n 点，再从 n 走到 1 点，不能走重复路，求最短路径

//可以看成从 1 走到 n 用两种路径，因为只能走一遍，所以边的容量为 1

所以建立一个附加的源点容量为2，费用为 0 到1点，附加的汇点容量为 2 ，费用为 0 到 n+1 点跑最小费用最大流即可

//# include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
# define eps 1e-8
# define INF 0x3f3f3f3f
# define pi  acos(-1.0)
# define MXN  1005
# define MXM  20050

struct Edge{
    int from, to, flow, cost, cap;
}edges[MXM*2];            //有向边数*2

struct MCMF{
    int n, m, idx;    //点，边，边数
    int flow, cost;
    vector<int> G[MXN];     //记录边
    int inq[MXN];           //BFS用
    int dis[MXN];           //层次
    int pre[MXN];           //上一条弧
    int adf[MXN];           //增量

    void Init(){
        idx=0;
        for (int i=0;i<=n+1;i++) G[i].clear();    //有附加点时要注意
    }

    void Addedge(int u,int v,int cost,int cap){
        edges[idx++] = (Edge){u,v,0,cost,cap};
        edges[idx++] = (Edge){v,u,0,-cost,0};
        G[u].push_back(idx-2);
        G[v].push_back(idx-1);
    }

    int Bellman(int s, int t)
    {
        memset(dis,0x3f,sizeof(dis));   //有附加点时要注意
        memset(inq,0,sizeof(inq));
        dis[s]=0, inq[s]=1, adf[s]=INF;
        queue<int> Q;
        Q.push(s);
        while (!Q.empty())
        {
            int u = Q.front(); Q.pop();
            inq[u]=0;
            for (int i=0;i<(int)G[u].size();i++)
            {
                Edge &e = edges[G[u][i]];
                if (dis[e.to] > dis[u] + e.cost && e.cap > e.flow)
                {
                    dis[e.to] = dis[u] + e.cost;
                    adf[e.to] = min(adf[u], e.cap-e.flow);
                    pre[e.to] = G[u][i];
                    if (!inq[e.to])
                    {
                        Q.push(e.to);
                        inq[e.to]=1;
                    }
                }
            }
        }
        if (dis[t]==INF) return false;

        flow+=adf[t];
        cost+=adf[t]*dis[t];
        int x=t;
        while(x!=s)
        {
            edges[pre[x]].flow+=adf[t];
            edges[pre[x]^1].flow-=adf[t];
            x=edges[pre[x]].from;
        }
        return true;
    }

    int MinCost(int s,int t)
    {
        flow = 0, cost = 0;
        while(Bellman(s, t));
        return cost;
    }
}F;

int main()
{
    scanf("%d%d",&F.n,&F.m);
    F.Init();
    for (int i=1;i<=F.m;i++)
    {
        int u,v,cost;
        scanf("%d%d%d",&u,&v,&cost);
        F.Addedge(u,v,cost,1);
        F.Addedge(v,u,cost,1);
    }
    F.Addedge(0, 1, 0, 2);
    F.Addedge(F.n, F.n+1, 0, 2);
    printf("%d
",F.MinCost(0,F.n+1));
    return 0;
}