【Floyed】【图论】村庄重建

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思路：

这题n比较小，floyed可以水过去，
其实就接近裸的floyed吧
边我用的邻接矩阵存的

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$Code$:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int f[250][250],a[250];
int n,m;
void floyd (int k)//floyed
{
    
    
	for (int i = 0; i < n ;++i)
	  for (int j = 0; j < n ;++j)
	  if(f[i][j]>f[i][k]+f[k][j])
	  f[i][j] = f[j][i] = f[i][k] + f[k][j]; 
}
int main()
{
    
    
	freopen ("rebuild.in","r",stdin);
	freopen ("rebuild.out","w",stdout);
    scanf ("%d%d", &n,&m);
    for (int i = 0; i < n; ++i)
	    scanf ("%d",&a[i]);
	for (int i = 0; i < n ;++i)
	    for (int j = 0; j < n ;++j)
	    f[i][j] = 1e9;
	for (int i = 0; i < n ;++i)
	    f[i][i] = 0;
	    int u,v,w;
	for (int i = 1; i <= m; ++i)
	{
    
    
		scanf("%d%d%d",&u,&v,&w);
		f[u][v] = f[v][u] = w;//邻接矩阵
	}
	int Q;

	scanf ("%d", &Q);//询问次数
	int p = 0;
	for (int i = 1; i <= Q; ++i)
	{
    
    
		scanf("%d%d%d",&u,&v,&w);
		while(a[p] <= w && p < n)  
		{
    
    
			floyd (p);
			p++;
		}  
		if(a[u] > w || a[v] > w) printf ("-1\n");  else
		{
    
    
			if(f[u][v] == 1e9) printf ("-1\n");  else
				printf("%d\n",f[u][v]);
		} 
	} 
    
	return 0;
}