Simple application of parallel search set (red alarm)

https://pintia.cn/problem-sets/994805046380707840/problems/994805063963230208

Ideas

The meaning of this code is to divide the graph into blocks , The nodes that can go through are merged into a set , The total number of blocks is num1. Then remove a node , Block again , The total number of blocks is num2. If num2-1（ Remove a single node ） And num1 Equal or num2（ After removing this node , The block where the node is located is still interconnected ） be equal to num1, Connectivity is not affected .

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
int father[550],road[5005][5],over,vis[5005],check[5005];
int find(int x){
    
	while(father[x]!=x){
    
		x=father[x];
		}
		return x;
	}
void conbine(int x,int y){
    
	if(find(x)!=find(y)){
    
		father[find(x)]=find(y);
		} 
	}
int main(){
    
	int n,m,k,a,b;
	cin >> n >>m;
	for(int i = 0;i <= n;i++){
    
		father[i] = i;
		}
	for(int i = 1;i <= m;i++){
    
		cin >> a >> b;
		road[i][1] = a;
		road[i][2] = b;
		conbine(a,b);
		}
	int num1 = 0;
	for(int i = 0;i < n;i++){
    
		if(father[i]==i)num1++;
		}
	cin >> k;
	// Yes K To deal with 
	for(int i = 1;i<= k;i++){
    
		int num2 = 0;
		cin >> a;
		check[a]=1;
		if(over)continue;
		for(int j = 0;j<n;j++){
    
			father[j]=j;
			}
		for(int j = 1;j <= m;j++){
    
			if(road[j][1]==a||road[j][2]==a||vis[j]){
    
				vis[j]=1;
				continue;
			}
			conbine(road[j][1],road[j][2]);
			}
			for(int j = 0;j < n;j++){
    
				if(j==father[j])num2++;
				}
			//cout<< num1 << " " << num2 <<endl;
			if(num1==num2-1||num1==num2){
    
			printf("City %d is lost.\n",a);
				}
			else if(num2-1 > num1){
    
				printf("Red Alert: City %d is lost!\n",a);
				}
			num1 = num2;
				over = 1;
			for(int j = 0;j < n;j++){
    
				if(!check[j])over = 0;
				}
				if(over)puts("Game Over.");
		}
	return 0;
	}