L2-013 Red Alert (25 points)

L2-013 Red Alert (25 points)

War to maintain connectivity between various cities is very important. This question requires you to write a program alarm, when the city lost a lead country is split into multiple zones can not communicate, it emits red alert. Note: If the country should not have full connectivity, k is divided regions, a lost city does not alter the connectivity between other cities, do not alarm.

Input formats:

Input two integers are given in the first row N(0 <  N ≤ 500) and M(≤ 5000), respectively, the number of city (city then default from 0 to N-1 numbered) and connections of the two cities via strip. Subsequently Mrows each given two cities connected via a number, separated by a space therebetween. It gives information was captured after city information, that is a positive integer Kand subsequently Ka number of cities were captured.

Note: Enter the city was captured guarantee given number is legitimate and no repeat, but does not guarantee passage given no repeat.

Output formats:

Each city was captured, if it will change the connectivity of the entire country, the output Red Alert: City k is lost!, which kis the city's number; otherwise, only the output City k is lost.can be. If the country lost its last city, then add a line output Game Over..

Sample input:

5 4
0 1
1 3
3 0
0 4
5
1 2 0 4 3

Sample output:

City 1 is lost.
City 2 is lost.
Red Alert: City 0 is lost!
City 4 is lost.
City 3 is lost.
Game Over.

 

#include<iostream>
using namespace std;
int   cnt ,n,mp[509][509] , vis[509] , v[509];
void  dfs( int node ){
	  for( int i=0;i<n;i++)
	       if( !v[i] && mp[node][i] ){
	       	   v[i] = 1;
	       	   dfs( i );
		   } 
}
int   check(   ){
	  for( int i=0 ; i<n ;i++)
	  	   v[i] = vis[i];
	  cnt =0; 
	  for( int j = 0 ;j<n;j++){ 
	       if( !v[j] ){ 
	 	        cnt++;
			    dfs( j ); 
	       }
	  }
      return cnt;
}
int   main(void){
      int m,a,b;
      cin>>n>>m;
      for( int i=1;i<=m;i++){
    	   cin>>a>>b;
    	   mp[a][b] = mp[b][a] =1;
	  }
	  int k;
	  cin>>k;
	  for( int i=1;i<=k;i++){
	       int cur = check( );   
	       int node;
	  	   cin >>node;
	       vis[node] = 1;
	       int  tmp = check( ); 
		   if(  cur >= tmp  )
	            cout<<"City "<<node<<" is lost."<<endl;
	       else {
		        cout<<"Red Alert: City "<<node<<" is lost!"<<endl;     
	       }
	       if( i == n )
              cout<<"Game Over."<<endl;
	  }
    
	
	  return 0; 
}