poj2135(84/600)

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

這玩意太簡單了不說了…

//#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; //需要拆點的話記得加倍
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];  //是否在隊列中
    int d[Vmax];    //Bellman-Ford
    int p[Vmax];    //上一條弧
    int a[Vmax];    //可改進量

    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; //鬆弛操作
                    p[e.to]=G[u][i];    //記錄上一條邊信息
                    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 不聯通，失敗退出
        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;
}