Distance opentrain

给定一个无向图以及其中的n个关键点，求其中最短的两个关键点的路径。
Dij，多源最短路。每个节点都记录最短路的起始点，如果两个不相同的起始点相连，那么这就是最短路。
注意用spfa会超时

#include <bits/stdc++.h>#include <vector>using namespace std;const int maxx = 1e9;const int maxn = 100010;const int mo = 50*maxn;struct Node {    int s, t;    Node() {t = maxx; s = 0;}    Node(int ss, int tt): s(ss), t(tt) {}    bool operator<(const Node &other) const {        return t < other.t;    }};Node dist[maxn];int d[mo], x, y, z, n, m, k, ans, l, r;bool flag[maxn], p[maxn];long long final;struct Edge {    int u, v, t;    Edge() {}    Edge(int uu, int vv, int tt): u(uu), v(vv), t(tt) {}    bool operator<(const Edge &other) const {        return t > other.t;    }};vector<Edge> G[maxn];priority_queue<Edge> q;void dij() {    int u, v;    for (int i = 1; i <= r; i++)        for (int j = 0; j < G[d[i]].size(); j++) {            q.push(G[d[i]][j]);        }    Edge e = q.top();    q.pop();    while (l < r) {        l++;        while (!q.empty() && dist[e.v].s != 0) {            if (dist[e.v].s != dist[e.u].s) {                ans = dist[e.u].t+dist[e.v].t+e.t;                break;            }            e = q.top();            q.pop();        }        if (q.empty() || ans != maxx) break;        u = e.u;        v = e.v;        dist[v].s = dist[u].s;        dist[v].t = dist[u].t+e.t;        d[++r] = v;        for (int i = 0; i < G[v].size(); i++) q.push(G[v][i]);    }}int main() {    //freopen("input.txt","r",stdin);    scanf("%d %d %d", &n, &k, &m);    r = 0;    ans = maxx;    for (int i = 1; i <= n; i++) {        scanf("%d", &x);        if (!flag[x]) {            flag[x] = true;            d[++r] = x;            dist[x] = Node(x, 0);        } else ans = 0;    }    if (ans == 0) {        printf("0\n");        return 0;    }    for (int i = 1; i <= m; i++) {        scanf("%d %d %d", &x, &y, &z);        G[x].push_back(Edge(x, y, z));        G[y].push_back(Edge(y, x, z));    }    dij();    final = ans;    final = final*3;    printf("%lld\n", final);}