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;
}