poj 2135 Farm Tour //最小费用流模板（dijstra实现）

最小费用流模板（dijstra实现）

Farm Tour

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 21105	Accepted: 8115

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：求不重复的两条最短路(无向边)。

可以转换为求1-n的两条没有公共边的路的最小花费，这样建立流量为2的最小费用流即可。（日常signed）

复杂度（FElogV)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<climits>
#include<cmath>
#include<vector>
#include<cstring>
#include<queue>
using namespace std;
#define int long long
const int max_n=1005;
struct no{int to,cap,cost,rev;};
struct qno{int v,d;}; //队列节点
bool operator<(const qno &x,const qno &y) {return x.d>y.d;} //队列cmp重写
int n,m;  //定点数 和边数
vector<no>g[max_n];
int dis[max_n],h[max_n]; //最短距离，顶点的势
int prevv[max_n],preve[max_n]; //在vector下记录路径

void addedge(int from,int to,int cap,int cost){
    g[from].push_back((no){to,cap,cost,g[to].size()});
    g[to].push_back((no){from,0,-cost,g[from].size()-1});
}
int min_cost_flow(int s,int t,int f){//注意n变化了！！！！！！！！！！！！！！！！！
    int res=0; //answer
    for(int i=0;i<=n;i++) h[i]=0; //初始化h
    while(f>0){
        priority_queue<qno>q;
        for(int i=0;i<=n;i++) dis[i]=INT_MAX/2;
        dis[s]=0;
        q.push((qno){s,0});
        while(!q.empty()){
            qno now=q.top();q.pop();
            int v=now.v;
            if(dis[v]<now.d) continue;
            for(int i=0;i<g[v].size();i++){
                no &e=g[v][i];
                if(e.cap>0&&dis[e.to]>dis[v]+e.cost+h[v]-h[e.to]){ //加入啦势
                    dis[e.to]=dis[v]+e.cost+h[v]-h[e.to];
                    prevv[e.to]=v;
                    preve[e.to]=i;
                    q.push((qno){e.to,dis[e.to]});
                }
            }
        }
        if(dis[t]==INT_MAX/2) return -1;
        for(int v=0;v<=n;v++) h[v]+=dis[v];
        int d=f; //找增广路
        for(int v=t;v!=s;v=prevv[v]) d=min(d,g[prevv[v]][preve[v]].cap);
        f-=d;res+=d*h[t];
        for(int v=t;v!=s;v=prevv[v]){
            no &e=g[prevv[v]][preve[v]];
            e.cap-=d;
            g[v][e.rev].cap+=d;
        }
    }
    return res;
}
signed main(){
    ios::sync_with_stdio(false);cin.tie(0);
    cin>>n>>m;
    for(int i=1;i<=m;i++){
        int u,v,w;cin>>u>>v>>w;
        addedge(u,v,1,w);
        addedge(v,u,1,w);
    }
    cout<<min_cost_flow(1,n,2);
    return 0;
}