POJ 2135.Farm Tour 最小费用流

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17307		Accepted: 6687

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

Sample Output

6


题目链接：http://poj.org/problem?id=2135
题意：有n块地，m条道路，问从1到n在返回1的最短距离，并且同一条道路不能经过2次。
思路：感觉像是先求出1-n的最短路，然后删除路径再次求最短路，但是这种方法,明显是错误的。将问题当作1-n的两条没有公共边的路径，将路径本身作为一条流量为1的路径，花费为路径长度，这样就不过是求流量为2的最小费用流

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
#define PI acos(-1.0)
const int maxn=3e5+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e18+7;
struct edge
{
    int from,to;
    int cap,cost;
    int rev;
};
int n;
vector<edge>G[maxn];
int h[maxn];
///顶点的势，取h(u)=(s到u的最短距离)，边e=(u,v)的长度变成d`(e)=d(e)+h(u)-h(v)>=0
int dist[maxn];
int prevv[maxn],preve[maxn];///前驱结点和对应的边
void addedge(int u,int v,int cap,int cost)
{
    edge e;
    e.from=u,e.to=v,e.cap=cap,e.cost=cost,e.rev=G[v].size();
    G[u].push_back(e);
    e.from=v,e.to=u,e.cap=0,e.cost=-cost,e.rev=G[u].size()-1;
    G[v].push_back(e);
}
int min_cost_flow(int s,int t,int f)
{
    int res=0;
    fill(h+1,h+n+1,0);
    while(f>0)
    {
        priority_queue<P,vector<P>,greater<P> >q;
        fill(dist+1,dist+n+1,inf);
        dist[s]=0;
        q.push(P(dist[s],s));
        while(!q.empty())
        {
            P p=q.top();
            q.pop();
            int u=p.second;
            if(dist[u]<p.first) continue;
            for(int i=0; i<G[u].size(); i++)
            {
                edge e=G[u][i];
                if(e.cap>0&&dist[e.to]>dist[u]+e.cost+h[u]-h[e.to])
                {
                    dist[e.to]=dist[u]+e.cost+h[u]-h[e.to];
                    prevv[e.to]=u;
                    preve[e.to]=i;
                    q.push(P(dist[e.to],e.to));
                }
            }
        }
        if(dist[t]==inf) return -1;
        for(int i=1; i<=n; i++) h[i]+=dist[i];
        int d=f;
        for(int i=t; i!=s; i=prevv[i])
            d=min(d,G[prevv[i]][preve[i]].cap);
        f-=d;
        res+=d*h[t];
        for(int i=t; i!=s; i=prevv[i])
        {
            edge &e=G[prevv[i]][preve[i]];
            e.cap-=d;
            G[i][e.rev].cap+=d;
        }
    }
    return res;
}
int main()
{
    int m;
    scanf("%d%d",&n,&m);
    for(int i=1; i<=m; i++)
    {
        int u,v,cost;
        scanf("%d%d%d",&u,&v,&cost);
        addedge(u,v,1,cost);
        addedge(v,u,1,cost);
    }
    int ans=min_cost_flow(1,n,2);
    cout<<ans<<endl;
    return 0;
}

最小费用流