解题思路

本题可以用最短路算法——Floyd

AC代码

#include<bits/stdc++.h>
#define inf 1e9
using namespace std;
const int N = 2e2 + 50;
int n, m, q, now = 0, a, b, c, t[N], G[N][N];int main()
{scanf("%d%d", &n, &m);for(int i = 0;i<n;i++)scanf("%d", &t[i]);for(int i = 0;i<n;i++){for(int j = 0;j<n;j++)G[i][j] = inf;G[i][i] = 0;}for(int i = 0;i<m;i++){scanf("%d%d%d", &a, &b, &c);G[a][b] = G[b][a] = c;}scanf("%d", &q);while(q--){scanf("%d%d%d", &a, &b, &c);while(t[now] <= c && now < n){for(int i = 0;i<n;i++)for(int j = 0;j<n;j++)if(G[i][j] > G[i][now] + G[now][j])G[i][j] = G[j][i] = G[i][now] + G[now][j];now++;}if(t[a] > c || t[b] > c)puts("-1");else{if(G[a][b] == inf) puts("-1");else printf("%d\n", G[a][b]);}}return 0;
}