Poj 2135 Farm Tour

http://poj.org/problem?id=2135

题意：约翰要带朋友参观农场。农场里有N块地，M条双向路径。第i条路连接ai和bi两块地，长度为ci。约翰住在一号地，他想带朋友们参观一圈，但每条路只能走一次。求路线的最小值。

题解：转换为求从1号到N号顶点的两条没有公共边的路径，就变成了最小费用流问题。因为涉及到往返，流量应该设为2。

求最小费用流的时候用Bellman_Ford求最短路效率要慢，改进后的Dijkstra要快。可是我交了之后发现Bellman_Ford用了32MS，Dijkstra要用47MS……

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <functional>
#include <set>
#include <fstream>

using namespace std;

const int maxn=10010;
const int INF=0x7fffffff;
int N,M;
int a[maxn],b[maxn],c[maxn];
struct edge {
    int to,cap,cost,rev;
};
int V;
vector<edge> G[maxn];
int dist[maxn];
int prevv[maxn],preve[maxn];
/*
void add_edge(int from,int to,int cap,int cost)
{
    G[from].push_back((edge){to,cap,cost,(int)G[to].size()});
    G[to].push_back((edge){from,0,-cost,(int)G[from].size()-1});
}

int min_cost_flow(int s,int t,int f)
{
    int res=0;
    while (f>0) {
        fill(dist, dist+V, INF);
        dist[s]=0;
        bool update=true;
        while (update) {
            update=false;
            for (int v=0; v<V; v++) {
                if (dist[v]==INF) {
                    continue;
                }
                for (int i=0; i<G[v].size(); i++) {
                    edge &e=G[v][i];
                    if (e.cap>0&&dist[e.to]>dist[v]+e.cost) {
                        dist[e.to]=dist[v]+e.cost;
                        prevv[e.to]=v;
                        preve[e.to]=i;
                        update=true;
                    }
                }
            }
        }
        if (dist[t]==INF) {
            return -1;
        }
        int d=f;
        for (int v=t; v!=s; v=prevv[v]) {
            d=min(d,G[prevv[v]][preve[v]].cap);
        }
        f-=d;
        res+=d*dist[t];
        for (int v=t; v!=s; v=prevv[v]) {
            edge &e=G[prevv[v]][preve[v]];
            e.cap-=d;
            G[v][e.rev].cap+=d;
        }
    }
    return res;
}
*/
int h[maxn];
typedef pair<int, int> P;
void add_edge(int from,int to,int cap,int cost)
{
    G[from].push_back((edge){to,cap,cost,(int)G[to].size()});
    G[to].push_back((edge){from,0,-cost,(int)G[from].size()-1});
}

int min_cost_flow(int s,int t,int f)
{
    int res=0;
    fill(h, h+V, 0);
    while (f>0) {
        priority_queue<P,vector<P>,greater<P> > que;
        fill(dist, dist+V, INF);
        dist[s]=0;
        que.push(P(0,s));
        while (!que.empty()) {
            P p=que.top();
            que.pop();
            int v=p.second;
            if (dist[v]<p.first) {
                continue;
            }
            for (int i=0; i<G[v].size(); i++) {
                edge &e=G[v][i];
                if (e.cap>0&&dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]) {
                    dist[e.to]=dist[v]+e.cost+h[v]-h[e.to];
                    prevv[e.to]=v;
                    preve[e.to]=i;
                    que.push(P(dist[e.to],e.to));
                }
            }
        }
        if (dist[t]==INF) {
            return -1;
        }
        for (int v=0; v<V; v++) {
            h[v]+=dist[v];
        }
        int d=f;
        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]) {
            edge &e=G[prevv[v]][preve[v]];
            e.cap-=d;
            G[v][e.rev].cap+=d;
        }
    }
    return res;
}

int main()
{
    //freopen("/Users/apple/Desktop/2135/2135/in", "r", stdin);
    //freopen("/Users/apple/Desktop/2135/2135/out", "w", stdout);
    while ((scanf("%d%d",&N,&M))!=EOF) {
        V=N;
        for (int i=0; i<M; i++) {
            scanf("%d%d%d",&a[i],&b[i],&c[i]);
        }
        int s=0,t=N-1;
        for (int i=0; i<M; i++) {
            add_edge(a[i]-1, b[i]-1, 1, c[i]);
            add_edge(b[i]-1, a[i]-1, 1, c[i]);
        }
        int res2=min_cost_flow(s, t, 2);
        printf("%d
",res2);
    }
    return 0;
}