已经按照题解的思路自己写了一份，但还是超时

按照floyd算法，每次用k去更新， 和模板不同的是， 这里的k并不是一直可以用，因此需要循环判断，用当前可以更新的点来更新最短距离

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;

const int N = 210, INF = 0x3f3f3f3f;
int d[N][N];
int t[N];
int n, m;

void floyd(int k)
{
    for (int i = 1; i <= n; i ++)
    {
        for (int j = 1; j <= n;j ++ )
        {
            d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
        }
    }
}

int main()
{
    cin >> n >> m;

    for (int i = 1; i <= n; i ++ ) scanf("%d", &t[i]);

    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++)
            if (i == j) d[i][j] = 0;
            else d[i][j] = INF;

    while (m -- )
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        a ++, b ++;
        d[a][b] = d[b][a] = c;
    }

    int q;
    cin >> q;

    while (q -- )
    {
        int x, y, time;

        scanf("%d%d%d", &x, &y, &time);
        x++, y ++;
        int current = 1;
        while (t[current] <= time && current <= n)
        {
            floyd(current);
            current ++;
        }

        if (t[x] > time || t[y] > time || d[x][y] >= INF / 2) cout << -1 << endl;
        else cout << d[x][y] << endl;
    }

    return 0;
}