求单源最短路使用Dijkstra算法，堆优化版本时间复杂度在O（nlogn）
题目链接：https://www.luogu.com.cn/record/153709192（注意此图是有向图）
使用优先队列使每次出队的都是距离最小值对应点

#include <iostream>
#include <queue>
#include <cstring>
#include <vector>
using namespace std;

const int N = 1e6 + 10, M = 1e9 + 10;
int e[N * 2], ne[N * 2], h[N];
long long w[2 * N], idx;
long long dist[N];
int n, m, s;
bool st[N];
typedef pair<int, int> pii;

void add(int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c,
        h[a] = idx++;
}

void dijkstra() {
    memset(dist, M, sizeof dist);
    dist[s] = 0;
    priority_queue<pii, vector<pii>, greater<pii>> heap;
    heap.push({ 0, s });
    while (heap.size()) {
        pii t = heap.top();
        heap.pop();
        int ver = t.second, distance = t.first;
        if (st[ver])
            continue;
        st[ver] = true;
        for (int i = h[ver]; i != -1; i = ne[i]) {
            int j = e[i];
            if (dist[j] > distance + w[i]) {
                dist[j] = distance + w[i];
                heap.push({dist[j],j});
            }
        }
    }
}

int main() {
    cin >> n >> m >> s;
    int a, b, c;
    memset(h, -1, sizeof h);
    for (int i = 0; i < m; i++) {
        cin >> a >> b >> c;
        add(a, b, c);
    }
    dijkstra();
    for (int i = 1; i <= n; i++)
        cout << dist[i] << ' ';
    return 0;
}