Dijkstra with priority queue

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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;}
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