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最新推荐文章于 2020-09-13 23:41:39 发布

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FJ在农场展示路径，他的农场有N个字段，用M条路径连接。他想找到一条从房子出发，经过一些字段，最后到达谷仓的最短路线，且返回时不能重复走同一路径。计算出这样的最短路线长度。

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

这玩意太简单了不说了…

//#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
int INF=0x3f3f3f3f;
const int Vmax=2005; //需要拆点的话记得加倍
namespace MCMF{
    struct Edge{
        int from,to,cap,flow,cost;
        Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w){}
    };

    int n,m;
    vector<Edge>edges;
    vector<int>G[Vmax];
    int inq[Vmax];  //是否在队列中
    int d[Vmax];    //Bellman-Ford
    int p[Vmax];    //上一条弧
    int a[Vmax];    //可改进量

    void init(int _Vsz){
        n=_Vsz;
        for(int i=0;i<=n;i++) G[i].clear();
        edges.clear();
    }

    void adde(int from,int to,int cap,int cost){
        edges.push_back(Edge(from, to, cap, 0, cost));
        edges.push_back(Edge(to, from, 0, 0, -cost));
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool SPFA(int s,int t,int& flow,long long& cost){
        for(int i=0;i<=n;i++) d[i]=INF;
        memset(inq, 0, sizeof(inq));
        d[s]=0;
        inq[s]=1;
        p[s]=0;
        a[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<G[u].size();i++){
                Edge& e=edges[G[u][i]];
                if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
                    d[e.to]=d[u]+e.cost; //松弛操作
                    p[e.to]=G[u][i];    //记录上一条边信息
                    a[e.to]=min(a[u], e.cap-e.flow);
                    if(!inq[e.to]){
                        q.push(e.to);
                        inq[e.to]=1;
                    }
                }
            }
        }
        if(d[t]==INF) return false; //s-t 不联通，失败退出
        flow+=a[t];
        cost+=(long long)d[t]*(long long)a[t];
        for(int u=t;u!=s;u=edges[p[u]].from){
            edges[p[u]].flow+=a[t];
            edges[p[u]^1].flow-=a[t];
        }
        return true;
    }

    int MincostMaxflow(int s,int t,long long& cost){
        int flow=0;
        cost=0;
        while(SPFA(s, t, flow, cost));
        return flow;
    }
}
int main()
{
    int n,m,q,w,e;
    cin>>n>>m;
    MCMF::init(n+10);
    MCMF::adde(0,1,2,0);
    for(int a=1;a<=m;a++)
    {
        scanf("%d%d%d",&q,&w,&e);
        MCMF::adde(q,w,1,e);
        MCMF::adde(w,q,1,e);
    }
    int st=0,ed=n+1;
    MCMF::adde(n,n+1,100,0);
    long long minCost, maxFlow;
    maxFlow = MCMF::MincostMaxflow(st, ed, minCost);
    cout<<minCost<<endl;
}