poj 1511 正向 反向 构两个图

有向图 源点为1 求源点到其他各点的最短距离之和 再在其他点到源点的最短距离之和 再加起来 多源点一终点 只要反向构图就行了

Sample Input

2 //T
2 2 //结点数 边数
1 2 13 //u v w
2 1 33
4 6
1 2 10
2 1 60
1 3 20
3 4 10
2 4 5
4 1 50
Sample Output

46
210

堆优化：  跑了6S...

 # include <iostream>
 # include <cstdio>
 # include <cstring>
 # include <algorithm>
 # include <cmath>
 # include <queue>
 # define LL long long
 using namespace std ;

 const int INF=0x3f3f3f3f;
 const int MAXN=;
 struct qnode
 {
     int v;
     int c;
     qnode(int _v=,int _c=):v(_v),c(_c){}
     bool operator <(const qnode &r)const
     {
         return c>r.c;
     }
 };
 struct Edge
 {
     int v,cost;
     Edge(int _v=,int _cost=):v(_v),cost(_cost){}
 };
 vector<Edge>E[MAXN];
 bool vis[MAXN];
 int dist[MAXN];
 int u[MAXN] , v[MAXN] , w[MAXN] ;
 int n ;
 void Dijkstra(int start)//点的编号从1开始
 {
     memset(vis,false,sizeof(vis));
     for(int i=;i<=n;i++)dist[i]=INF;
     priority_queue<qnode>que;
     while(!que.empty())que.pop();
     dist[start]=;
     que.push(qnode(start,));
     qnode tmp;
     while(!que.empty())
     {
         tmp=que.top();
         que.pop();
         int u=tmp.v;
         if(vis[u])continue;
         vis[u]=true;
         for(int i=;i<E[u].size();i++)
         {
             int v=E[tmp.v][i].v;
             int cost=E[u][i].cost;
             if(!vis[v]&&dist[v]>dist[u]+cost)
             {
                 dist[v]=dist[u]+cost;
                 que.push(qnode(v,dist[v]));
             }
         }
     }
 }
 void addedge(int u,int v,int w)
 {
     E[u].push_back(Edge(v,w));
 }

 int main ()
 {
    // freopen("in.txt","r",stdin) ;
     int m ;
     int T ;
     scanf("%d" , &T) ;
     while (T--)
     {
         scanf("%d %d" , &n , &m) ;
         LL ans =  ;
         int i , j ;
         for(i=;i<=m;i++)
             scanf("%d%d%d" , &u[i] , &v[i] , &w[i]) ;

         for(i=;i<=n;i++)
             E[i].clear();
         for(i=;i<=m;i++)
             addedge(u[i],v[i],w[i]) ;
         Dijkstra() ;
         for(i=;i<=n;i++)
             ans += dist[i] ;

         for(i=;i<=n;i++)
             E[i].clear();
         for(i=;i<=m;i++)
             addedge(v[i],u[i],w[i]) ;
         Dijkstra() ;
         for(i=;i<=n;i++)
             ans += dist[i] ;

         cout<<ans<<endl ;

     }

     return  ;
 }