Dijkstra with priority queue

POJ 1511 Invitation Cards（单源最短路，优先队列优化的Dijkstra）

//============================================================================
// Name        : POJ.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
/*
 * 使用优先队列优化Dijkstra算法
 * 复杂度O(ElogE)
 * 注意对vector<Edge>E[MAXN]进行初始化后加边
 */
const int INF=0x3f3f3f3f;
const int MAXN=1000010;
struct qnode
{
    int v;
    int c;
    qnode(int _v=0,int _c=0):v(_v),c(_c){}
    bool operator <(const qnode &r)const
    {
        return c>r.c;
    }
};
struct Edge
{
    int v,cost;
    Edge(int _v=0,int _cost=0):v(_v),cost(_cost){}
};
vector<Edge>E[MAXN];
bool vis[MAXN];
int dist[MAXN];
void Dijkstra(int n,int start)//点的编号从1开始
{
    memset(vis,false,sizeof(vis));
    for(int i=1;i<=n;i++)dist[i]=INF;
    priority_queue<qnode>que;
    while(!que.empty())que.pop();
    dist[start]=0;
    que.push(qnode(start,0));
    qnode tmp;
    while(!que.empty())
    {
        tmp=que.top();
        que.pop();
        int u=tmp.v;
        if(vis[u])continue;
        vis[u]=true;
        for(int i=0;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 A[MAXN],B[MAXN],C[MAXN];
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n,m;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
            scanf("%d%d%d",&A[i],&B[i],&C[i]);
        for(int i=1;i<=n;i++)E[i].clear();
        for(int i=0;i<m;i++)addedge(A[i],B[i],C[i]);
        Dijkstra(n,1);
        long long ans=0;
        for(int i=1;i<=n;i++)ans+=dist[i];
        for(int i=1;i<=n;i++)E[i].clear();
        for(int i=0;i<m;i++)addedge(B[i],A[i],C[i]);
        Dijkstra(n,1);
        for(int i=1;i<=n;i++)ans+=dist[i];
        printf("%I64d\n",ans);
    }
    return 0;
}