poj-1511
从1节点到所有节点的最短路和,加上所有节点返回1节点的最短路和,刚开始的方法时间复杂度有毒啊
其实只要把边全反向重装一次就好了哈哈哈
好了就是这样,套路了一个dijkstra+优先队列
#include<cstdio> #include<queue> #include<iostream> #include<cstring> #include<vector> using namespace std; const int INF = 0x3f3f3f3f; const int maxn = 1000000 + 100; typedef long long ll; int n, m; int dis[maxn]; int vis[maxn]; int head[maxn]; struct Node { int to; int len; int next; }G[maxn]; bool operator < (const Node a, const Node b) { if (a.len > b.len) return true; else return false; } int cnt = 1; void insert(int be, int en, int len) { G[cnt].to = en; G[cnt].next = head[be]; head[be] = cnt; G[cnt].len = len;//头插法 cnt++; } int dijstra(int be) { memset(vis, 0, sizeof(vis)); for (int i = 0; i <= n; i++) dis[i] = INF; dis[be] = 0; priority_queue<Node>que; Node a; a.to = be; a.len = 0; que.push(a); while (!que.empty()) { Node ans = que.top(); que.pop(); if (vis[ans.to] == 0) { vis[ans.to] = 1; for (int i = head[ans.to]; i; i = G[i].next) { int p = G[i].to; if (!vis[p] && dis[p] > dis[ans.to] + G[i].len) { dis[p] = dis[ans.to] + G[i].len; Node ac; ac.len = dis[p]; ac.to = p; que.push(ac); } } } } return 0; } int T; int list_be[maxn]; int list_en[maxn]; int list_val[maxn]; int main() { scanf("%d", &T); while (T--) { scanf("%d %d", &n, &m); memset(head, 0, sizeof(head)); cnt = 1; int a, b, c; for (int i = 0; i < m; i++) { scanf("%d %d %d", &a, &b, &c); list_be[i] = a; list_en[i] = b; list_val[i] = c; insert(a, b, c); } dijstra(1); ll ans = 0; for (int i = 1; i <= n; i++) { ans += dis[i]; } memset(head, 0, sizeof(head)); cnt = 1; for (int i = 0; i < m; i++) { insert(list_en[i], list_be[i], list_val[i]); } dijstra(1); for (int i = 1; i <= n; i++) { ans += dis[i]; } printf("%lld\n", ans); } return 0; }
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