AcWing 1125. Cattle travel problem solution (shortest path, diameter)

AcWing 1125. The journey of cattle
What is more troublesome is the calculation of the diameter and the diameter after connecting the points of two different pastures ,floyd The necessity of seeking the shortest path is not very strong

#include<bits/stdc++.h>

using namespace std;

#define x first
#define y second
#define db double

typedef pair<db, db>PDD;

const int N = 200;
const double MNF = 1e20;

int n, m;
char g[N][N];  // Storage matrix  
PDD q[N];  // Store coordinates  
db d[N][N];  // Store the shortest distance 
db maxd[N];  // The longest distance to store in the same pasture 

db get_dist(PDD a, PDD b){
      // Find the distance between two points  
	db dx = a.x - b.x, dy = a.y - b.y;
	return sqrt(dx * dx + dy * dy);
}

int main()
{
    
	cin>>n;
	for(int i = 0; i < n; i ++ ){
    
		cin>>q[i].x>>q[i].y;
	}
	
	for(int i = 0; i < n; i ++ ) cin>>g[i]; 
	
	for(int i = 0; i < n; i ++ ){
    
		for(int j = 0; j < n; j ++ ){
    
			if(i == j) d[i][j] = 0;
			else if(g[i][j] == '1') d[i][j] = get_dist(q[i], q[j]);
			else d[i][j] = MNF;
		}
	}
	
	//Floyd Find the shortest path  
	for(int k = 0; k < n; k ++ ){
    
		for(int i = 0; i < n; i ++ ){
    
			for(int j = 0; j < n; j ++ ){
    
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
			}
		}
	}
	
	db res1 = 0;
	for(int i = 0; i < n ; i ++ ){
    
		for(int j = 0; j < n; j ++ ){
    
			if(d[i][j] < MNF / 2){
    
				// Here I'm wondering why I can't use the matrix 1 Judge two points that are not connected 
				// The reason is that the number of matrices of two indirectly connected points in the same pasture is also 1, Therefore, the value of the matrix cannot be used to judge whether two points are a pasture 
				maxd[i] = max(d[i][j], maxd[i]);  // Find the diameter of each point 
			} 
		} 
		res1 = max(maxd[i], res1);
	}
	
	db res = MNF;  // initialization 
	for(int i = 0; i < n; i ++ ){
    
		for(int j = 0; j < n; j ++ ){
    
			if(d[i][j] > MNF / 2){
      // Here I wonder why I can't use the 0 Judge two points that are not connected 
			// The reason is that the matrix number of two points that are not directly connected in the same pasture is also 0, Therefore, it is impossible to judge whether two points are a pasture with a matrix  
				res = min(res, maxd[i] + maxd[j] + get_dist(q[i], q[j]));
			}
		}
	} 
	
	printf("%.6lf\n", max(res, res1));
	return 0;
}