POJ2135：Farm Tour

题意：给定一个无向图，从1走到n再从n走回1，每个边只能走一遍，求最短路

题解：可以定义一个源点s，和一个汇点t

s和1相连容量为2，费用为0，

t和n相连容量为2，费用为0

然后所用的边的容量都定为1，跑一遍最小费用最大流即可

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
#define MAXN 1000+10
#define INF 0x7f7f7f7f
#define ll long long
using namespace std;
struct Edge{
    int from,to,cap,flow,cost;
    Edge(int u=0,int v=0,int c=0,int f=0,int w=0){
        from=u,to=v,cap=c,flow=f,cost=w;
    }
};
int n,m;
vector<Edge> edges;
vector<int> G[MAXN];
int d[MAXN];
int b[MAXN];
int a[MAXN];
int p[MAXN];
void AddEdge(int u,int v,int cap,int cost){
    edges.push_back(Edge(u,v,cap,0,cost));
    edges.push_back(Edge(v,u,0,0,-cost));
    int t=edges.size();
    G[u].push_back(t-2);
    G[v].push_back(t-1);
}
int SPFA(int s,int t,int &flow,ll &cost){
    memset(d,0x7f,sizeof(d));
    memset(b,0,sizeof(b));

    queue<int> q;
    q.push(s);
    b[s]=1;
    d[s]=0;
    a[s]=INF;
    p[s]=0;

    while(!q.empty()){
        int x=q.front(); q.pop();
        b[x]=0;
        for(int i=0;i<G[x].size();i++){
            Edge& e=edges[G[x][i]];
            if(e.cap>e.flow&&d[e.to]>d[x]+e.cost){
                d[e.to]=d[x]+e.cost;
                a[e.to]=min(a[x],e.cap-e.flow);
                p[e.to]=G[x][i];
                if(!b[e.to]){
                    b[e.to]=1;
                    q.push(e.to);
                }
            }
        }
    }
    if(d[t]==INF){
        return 0;
    }    

    flow+=a[t];
    cost+=1LL*a[t]*d[t];
    for(int x=t;x!=s;x=edges[p[x]].from){
        edges[p[x]].flow+=a[t];
        edges[p[x]^1].flow-=a[t];
    }
    return 1;
}
ll MincostMaxflow(int s,int t){
    int flow=0;
    ll cost=0;
    while(SPFA(s,t,flow,cost));
    return cost;
}
void solve(){
    printf("%lld
",MincostMaxflow(1,n+2));
}
void init(){
    memset(a,0,sizeof(a));
    memset(p,0,sizeof(p));
    edges.clear();
    for(int i=0;i<=n;i++){
        G[i].clear();
    }    
    for(int i=1;i<=m;i++){
        int u,v,w;scanf("%d%d%d",&u,&v,&w);
        u++,v++;
        AddEdge(u,v,1,w);
        AddEdge(v,u,1,w);
    }
    AddEdge(1,2,2,0);
    AddEdge(n+1,n+2,2,0);
}
int main()
{
    while(~scanf("%d%d",&n,&m)){
        init();
        solve();
    }
    return 0;
}