P1119 灾后重建

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#include<bits/stdc++.h>
using namespace std;

const int MAXN = 205;
const int INF = 1e9; // 定义一个足够大的数代表无穷大

int n, m;
int t[MAXN];          //存每个村庄的修复时间
int dist[MAXN][MAXN]; 

int main() {
 
    ios::sync_with_stdio(false);
    cin.tie(0);

    
    cin >> n >> m;

    
    for (int i = 0; i < n; i++) {
        cin >> t[i];
    }

    
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (i == j) dist[i][j] = 0;
            else dist[i][j] = INF;
        }
    }

   
    for (int i = 0; i < m; i++) {
        int u, v, w;
        cin >> u >> v >> w;
       
        dist[u][v] = dist[v][u] = w; 
    }

    
    int q;
    cin >> q;

    int now = 0; //记录当前floyd解锁到了哪个村庄了

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

       //按时间决定要遍历作为中转点的村庄数量，这样相当于只跑了now层，等程序结束也才跑了一个三重循环
        while (now < n && t[now] <= time) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
            
                    if (dist[i][now] != INF && dist[now][j] != INF) {
                        dist[i][j] = min(dist[i][j], dist[i][now] + dist[now][j]);
                    }
                }
            }
            
            now++; 
        }

        //没修好不可达，不连通不可达
        if (t[x] > time || t[y] > time || dist[x][y] == INF) {
            cout << -1 << endl;
        } else {
            cout << dist[x][y] << endl;
        }
    }

    return 0;
}