标签:blog io for ar div amp log size
题意:给你n个点的有向图,从1点到其他所有点又从其他点回到1点的最短路。
思路:可以求一次从1点出发的最短路,再反向建图,再求一次从1出发的最短路,把两次的结果加起来就是题目所求。由于边比较多,
所以最好用Dijkstra+优先队列,或者SPFA;
#include<cstdio> #include<cstring> #include<queue> #include<algorithm> const int maxn = 1000009; const int inf = 1<<30; #include<vector> struct node { int v,w; node(int v,int w):v(v),w(w){} bool operator < (const node a)const { return w > a.w; } }; std::vector<node> eg[maxn]; std::priority_queue<node> q; int n,m; int vis[maxn]; int dis[maxn]; int a[maxn],b[maxn],c[maxn]; void DIJ() { int i; memset(vis,0,sizeof(vis)); for(i=0;i<=n;i++) dis[i] = inf; while(!q.empty())q.pop(); dis[1] = 0; vis[1] = 0; int size = eg[1].size(); for(i=0; i < size; i++) { q.push(node(eg[1][i].v,eg[1][i].w)); dis[eg[1][i].v] = eg[1][i].w; } while(!q.empty()) { int u = q.top().v; int w = q.top().w; q.pop(); if(!vis[u]) { vis[u] = 1; size = eg[u].size(); for(i=0; i<size; i++) { int v = eg[u][i].v; int w = eg[u][i].w; if(!vis[v] && dis[u] + w < dis[v]) { dis[v] = dis[u] + w; q.push(node(v,dis[v])); } } } } } void work() { int i,sum = 0; DIJ(); for(i=2; i<=n; i++) sum += dis[i]; //printf("%d ",dis[i]);} for(i=0; i<=n; i++) eg[i].clear(); for(i=0;i<m;i++) eg[b[i]].push_back(node(a[i],c[i])); // printf("\n"); DIJ(); for(i=2; i <= n;i++) sum += dis[i]; //printf("%d ",dis[i]);} printf("%d\n",sum); } void input() { int i; scanf("%d%d",&n,&m); for(i=0 ;i < m; i++) scanf("%d%d%d",&a[i],&b[i],&c[i]); for(i=0;i<=n;i++) eg[i].clear(); for(i=0;i<m;i++) eg[a[i]].push_back(node(b[i],c[i])); } int main() { int t; scanf("%d",&t); while(t--) { input(); work(); } return 0; } /* 3 2 2 1 2 13 2 1 33 4 6 1 2 10 2 1 60 1 3 20 3 4 10 2 4 5 4 1 50 2 2 1 2 13 2 1 33 */
标签:blog io for ar div amp log size
原文地址:http://www.cnblogs.com/BruceNoOne/p/3900518.html