7-3 Telefraud Detection

7-3 Telefraud Detection  

Telefraud（电信诈骗） remains a common and persistent problem in our society. In some cases, unsuspecting victims lose their entire life savings. To stop this crime, you are supposed to write a program to detect those suspects from a huge amount of phone call records.

A person must be detected as a suspect if he/she makes more than K short phone calls to different people everyday, but no more than 20% of these people would call back. And more, if two suspects are calling each other, we say they might belong to the same gang. A makes a short phone call to B means that the total duration of the calls from A to B is no more than 5 minutes.

Input Specification:

Each input file contains one test case. For each case, the first line gives 3 positive integers K (≤500, the threshold（阈值） of the amount of short phone calls), N (≤10​3​​, the number of different phone numbers), and M (≤10​5​​, the number of phone call records). Then M lines of one day's records are given, each in the format:

caller receiver duration

where caller and receiver are numbered from 1 to N, and duration is no more than 1440 minutes in a day.

Output Specification:

Print in each line all the detected suspects in a gang, in ascending order of their numbers. The gangs are printed in ascending order of their first members. The numbers in a line must be separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.

If no one is detected, output None instead.

Sample Input 1:

5 15 31
1 4 2
1 5 2
1 5 4
1 7 5
1 8 3
1 9 1
1 6 5
1 15 2
1 15 5
3 2 2
3 5 15
3 13 1
3 12 1
3 14 1
3 10 2
3 11 5
5 2 1
5 3 10
5 1 1
5 7 2
5 6 1
5 13 4
5 15 1
11 10 5
12 14 1
6 1 1
6 9 2
6 10 5
6 11 2
6 12 1
6 13 1

Sample Output 1:

3 5
6

Note: In sample 1, although 1 had 9 records, but there were 7 distinct receivers, among which 5 and 15 both had conversations lasted more than 5 minutes in total. Hence 1 had made 5 short phone calls and didn't exceed the threshold 5, and therefore is not a suspect.

Sample Input 2:

5 7 8
1 2 1
1 3 1
1 4 1
1 5 1
1 6 1
1 7 1
2 1 1
3 1 1

Sample Output 2:

None

#include
#include
using namespace std;
int mp1[1009][1009],vis[1009],father[1009];
int find( int x ){
	return father[x] == x ? x : find( father[x]);
}
void merge( int a ,int b ){
 	 int fa = find( a );
 	 int fb = find( b );
	 if( fa!=fb  ){
	 	 father[fa] = fb;
	 } 
}
int main(void){
	int k,n,m;
	scanf("%d%d%d",&k,&n,&m);
	for( int i=1;i<=m;i++){
	     int c,r ,d;
	     scanf("%d%d%d",&c,&r,&d);
	     mp1[c][r]+=d;    
	} 
	set s;
	for( int i=1;i<=n;i++){
		 int cnt1= 0 ,cnt2=0; 
		 for( int j=1;j<=n;j++){ 
		    if( mp1[i][j]<=5 && mp1[i][j]!=0 ){ 
		        cnt1++;
		        if( mp1[j][i]   )
				    cnt2++;    
		    } 
		 }  
		 if( cnt1 >k )
		 if( (double)cnt1*0.2 >= (double)cnt2 ) 
		     s.insert(i);    
	}
	for( int i=1;i<=n;i++)
	     father[i] = i;  
	for(  auto t:s){
		  for( int i=1;i<=n;i++){
		  	   if( s.find(i) !=s.end()  ){
		  	   	   if( mp1[t][i]  && mp1[i][t] ){
		  	   	       merge( t, i);
				   }
		       }
		  }
	} 
	bool is = false; 
	for( int i=1;i<=n;i++){ 
		 if( s.find(i) !=s.end() && !vis[i]){ 
		     vis[i] = 1;
		     printf("%d",i);
		     is = true;
		   for( int j=1;j<=n;j++){
		     if( s.find(j) !=s.end()&&!vis[j]){
		     	 if( find(i) == find(j)){
		     	     vis[j] =1;
					 printf(" %d",j);      
				 }
			 } 
		   } 
		   printf("\n");
	     }
	} 
	if( !is )
	  printf("None\n");
	return 0;
}