Farm Tour POJ - 2135 （最小费用流）

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include<queue>
using namespace std;

#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn =  1e4+5;
const int  mod = 1e9+7;

typedef  pair<int,int>P;
struct edge{
    int to,cap,cost,rev;
};

int V;
vector<edge> G[maxn];
int h[maxn];
int dist[maxn];
int prevv[maxn],preve[maxn];

void addedge(int from,int to,int cap,int cost)
{
    G[from].push_back((edge){to,cap,cost,G[to].size()});
    G[to].push_back((edge){from,0,-cost,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 N,M;
int a[maxn],b[maxn],c[maxn];

void solve()
{
    int s=0,t=N-1;
    V=N;
    for(int i=0;i<M;i++)
    {
        addedge(a[i]-1,b[i]-1,1,c[i]);
        addedge(b[i]-1,a[i]-1,1,c[i]);
    }
    printf("%d\n",min_cost_flow(s,t,2));
}

int main()
{
    cin>>N>>M;
    for(int i=0;i<M;i++)
    {
        scanf("%d%d%d",&a[i],&b[i],&c[i]);
    }
    solve();
}