给定一个有$n$个节点，$m$条边的无向联通图，有$k$个点有$portals$，当经过了某个点，如果这个点有$portal$，它就会永久开启，

对于任意两个开启的$portal$，我们可以不需要花时间穿行，问开启所有的$portal$需要多少时间。

可以考虑这是一个$1->S_v$，$S_v$是另$k$个$portals$联通的最小生成树，

因此我们的答案就是$1->nearest\in k$，再加上最小生成树的代价，

我们另这$k$个点去跑最短路，然后把原本的边{u,v,w}\{u, v, w\}{u,v,w}变为{nearest∈kofu,nearest∈kofv,dis[u]+dis[v]+w}\{nearest \in k\ of\ u, nearest \in k\ of\ v, dis[u] + dis[v] + w\}{nearest∈k of u,nearest∈k of v,dis[u]+dis[v]+w}

$k$点中与$u$最近的点，$k$点中与$v$最近的点，边权同时更新，最后只需要跑一个最小生成树计算代价即可。

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10;struct Res {int u, v;long long w;void read() {scanf("%d %d %lld", &u, &v, &w);}bool operator < (const Res &t) const {return w < t.w;}
}edge[N];struct Node {int u; long long w;bool operator < (const Node &t) const {return w > t.w;}
};int n, m, k, vis[N], fa[N], pre[N];long long dis[N];vector< pair<int, int> > G[N];int find(int rt) {return rt == fa[rt] ? rt : fa[rt] = find(fa[rt]);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d %d", &n, &m);for (int i = 1; i <= m; i++) {edge[i].read();G[edge[i].u].push_back({edge[i].v, edge[i].w});G[edge[i].v].push_back({edge[i].u, edge[i].w});}scanf("%d", &k);priority_queue<Node> q;memset(dis, 0x3f, sizeof dis);for (int i = 1, x; i <= k; i++) {scanf("%d", &x);q.push({x, 0});pre[x] = x;dis[x] = 0;}while (q.size()) {auto u = q.top().u;q.pop();if (vis[u]) {continue;}vis[u] = 1;for (auto &to : G[u]) {if (dis[to.first] > dis[u] + to.second) {dis[to.first] = dis[u] + to.second;pre[to.first] = pre[u];q.push({to.first, dis[to.first]});}}}long long ans = dis[1];for (int i = 1; i <= m; i++) {edge[i].w += dis[edge[i].u] + dis[edge[i].v];edge[i].u = pre[edge[i].u], edge[i].v = pre[edge[i].v];}sort(edge + 1, edge + 1 + m);for (int i = 1; i <= n; i++) {fa[i] = i;}for (int i = 1, cur = 1; i <= m && cur < n; i++) {int u = find(edge[i].u), v = find(edge[i].v);if (u ^ v) {cur++;fa[u] = v;ans += edge[i].w;}}printf("%lld\n", ans);return 0;
}