标签:print blank push nbsp priority efi href http for
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\n" , sum); } return 0; }
poj 1511 Invitation Cards(dijstra优化)
标签:print blank push nbsp priority efi href http for
原文地址:http://www.cnblogs.com/TnT2333333/p/6547917.html