poj 1511 Invitation Cards（dijstra优化）

题目链接：http://poj.org/problem?id=1511

题意：给出n个点和n条有向边，求所有点到源点1的来回最短路之和（保证每个点都可以往返源点1）

题目比较简单就是边和点的个数有点多所以可以用dijstra+优先队列这样复杂度就可以到v*logn

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#define inf 1000000000
using namespace std;
const int M = 1e6 + 10;
int n , m , a[M] , b[M] , c[M] , dis[M];
struct TnT {
    int v , w;
};
struct cmp {
    bool operator() (int x , int y) {
        return dis[x] > dis[y];
    }
};
vector<TnT>vc[M];
bool vis[M];
void dij(int s) {
    priority_queue<int , vector<int> , cmp>q;
    memset(vis , false , sizeof(vis));
    TnT gg;
    q.push(s);
    dis[s] = 0;
    while(!q.empty()) {
        int m = q.top();
        vis[m] = true;
        for(int i = 0 ; i < vc[m].size() ; i++) {
            gg = vc[m][i];
            if(dis[m] + gg.w < dis[gg.v]) {
                dis[gg.v] = dis[m] + gg.w;
                if(!vis[gg.v]) {
                    vis[gg.v] = true;
                    q.push(gg.v);
                }
            }
        }
        q.pop();
    }
}
int main() {
    int t;
    TnT gg;
    scanf("%d" , &t);
    while(t--) {
        scanf("%d%d" , &n , &m);
        for(int i = 1 ; i <= n ; i++) {
            vc[i].clear();
            dis[i] = inf;
        }
        for(int i = 1 ; i <= m ; i++) {
            scanf("%d%d%d" , &a[i] , &b[i] , &c[i]);
            gg.v = b[i] , gg.w = c[i];
            vc[a[i]].push_back(gg);
        }
        dij(1);
        long long sum = 0;
        for(int i = 1 ; i <= n ; i++) {
            sum += (long long)dis[i];
        }
        for(int i = 1 ; i <= n ; i++) {
            vc[i].clear();
            dis[i] = inf;
        }
        for(int i = 1 ; i <= m ; i++) {
            gg.v = a[i] , gg.w = c[i];
            vc[b[i]].push_back(gg);
        }
        dij(1);
        for(int i = 1 ; i <= n ; i++) {
            sum += (long long)dis[i];
        }
        printf("%lld
" , sum);
    }
    return 0;
    
}