J - Invitation Cards POJ - 1511

题意：共有 n 个站点，n 个志愿者从1 出发，分别到达不同的站点，晚上全部回到 1 ，求他们总的最小花费是多少？
思路：首先想到从 1 出发跑一遍 DJ，然后再以每一个点为起点出发跑 n-1 遍 DJ ,求出所有 dis[1] + dis2[i] 即可，但是看了一下复杂度， n <= 1e6，炸了
优化一下，从 1 出发跑一边 DJ 后，反向建图，然后再从 1 出发跑一遍，答案就有了
复杂度：O（2*nlogn）=4e7

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#define ll long long
using namespace std;
typedef pair<ll,int> PII;
const int maxn = 1e6+10;
int n, m;int u[maxn], v[maxn], w[maxn];inline int read() {char c = getchar(); int f = 1, x = 0;while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}return f * x;
}int head[maxn], cnt;
bool vis[maxn];
struct node{int to, next, w;
}e[maxn];void add(int u, int v, int w) {e[++cnt].next = head[u];e[cnt].to = v;e[cnt].w = w;head[u] = cnt;
}ll dis[maxn];void init() {memset(vis, 0, sizeof(vis));memset(head, 0, sizeof(head));cnt = 0;memset(dis, 0x3f3f3f3f3f3f3f3f, sizeof(dis));memset(e, 0, sizeof(e));
}ll dijkstra() { // 每次取出与连通块相邻的所有边中最小的一条加入连通图，直至所有边都在连通块内 dis[1] = 0;priority_queue<PII, vector<PII>, greater<PII> > heap;heap.push({0, 1});while(heap.size()) {PII now = heap.top(); heap.pop();ll dist = now.first; int u = now.second;if(vis[u]) continue; vis[u] = 1;for(int i = head[u]; i; i = e[i].next) {int to = e[i].to;if(dis[to] > dist + e[i].w) {dis[to] = dist +e[i].w; // 后面进入的点可以更新与其相邻的，先进入的点 heap.push({dis[to], to});}}}// 统计答案ll sum = 0;for(int i = 1; i <= n; i++) sum += dis[i];return sum;
}int main() {//  freopen("test.in", "r", stdin);int T = read();while(T--) {ll ans = 0;init();n = read(); m = read();for(int i = 1; i <= m; i++) {u[i] = read(), v[i] = read(), w[i] = read();add(u[i], v[i], w[i]);}       ans += dijkstra();init();for(int i = 1; i <= m; i++) add(v[i], u[i], w[i]);ans += dijkstra();printf("%lld\n", ans);}return 0;
}

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