
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;
}
原文:https://www.cnblogs.com/lesning/p/11329742.html