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=,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[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,,cost));
     edges.push_back(Edge(v,u,,,-cost));
     int t=edges.size();
     G[u].push_back(t-);
     G[v].push_back(t-);
 }
 int SPFA(int s,int t,int &flow,ll &cost){
     memset(d,0x7f,sizeof(d));
     memset(b,,sizeof(b));

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

     while(!q.empty()){
         int x=q.front(); q.pop();
         b[x]=;
         for(int i=;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]=;
                     q.push(e.to);
                 }
             }
         }
     }
     if(d[t]==INF){
         return ;
     }    

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