POJ_2135_Farm Tour(最小费用流)

Farm Tour

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13539		Accepted: 5138

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：给你一副无向图，问从1->n->1这样走一个来回所用的最短路径是多少，每条边只能走一次。

分析：最小费用流问题。把边的长度当成费用，每条边容量为1，由于是无向图，所以每条边要处理两次，即u->v，v->u都要加进去。把图建好后跑一遍流量为2的最小费用流得出最小费用即可。

题目链接：http://poj.org/problem?id=2135

代码清单：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;

#define end() return 0

typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;

const int maxn = 4000 + 5;
const int INF = 0x7f7f7f7f;

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){}
};

struct MCMF{
    int n,m;
    vector<Edge>edge; //边数的两倍
    vector<int>G[maxn]; //邻接表，G[i][j]表示i的第j条边在e数组中的序号
    int inq[maxn]; //是否在队列
    int d[maxn]; //Bellman-Ford
    int p[maxn]; //上一条弧
    int a[maxn]; //可改进量

    void init(int n){
        this -> n = n;
        for(int i=0;i<=n;i++) G[i].clear();
        edge.clear();
    }

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

    bool BellmanFord(int s,int t,int& flow,int& cost){
        memset(d,INF,sizeof(d));
        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=edge[G[u][i]];
                if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
                    d[e.to]=d[u]+e.cost;
                    p[e.to]=G[u][i];
                    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;
        flow+=a[t];
        cost+=d[t];
        if(flow==2) return false;
        for(int u=t;u!=s;u=edge[p[u]].from){
            edge[p[u]].flow+=a[t];
            edge[p[u]^1].flow-=a[t];
        }
        return true;
    }

    //需要保证初始网络中没有负权圈
    int MincostMaxflow(int s,int t){
        int flow=0,cost=0;
        while(BellmanFord(s,t,flow,cost));
        return cost;
    }
};

int N,M;
int a,b,c;
MCMF mcmf;

void input(){
    scanf("%d%d",&N,&M);
    mcmf.init(N);
    for(int i=0;i<M;i++){
        scanf("%d%d%d",&a,&b,&c);
        mcmf.addEdge(a,b,1,c);
        mcmf.addEdge(b,a,1,c);
    }
}

void solve(){
    printf("%d\n",mcmf.MincostMaxflow(1,N));
}

int main(){
    input();
    solve();
    end();
}