Telefraud（電信詐騙） 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 shortphone 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（閾值） of the amount of short phone calls), N (≤103, the number of different phone numbers), and M (≤105, 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<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<string>
#include<sstream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
#define inf 0x3f3f3f3f
#define LL long long
vector<int> sus;
int ma[1005][1005];
int fa[1005];
int find(int x) //查找
{
int r=x,i=x,j;
while(fa[r]!=r)
r=fa[r];
while(fa[i]!=i)
j=fa[i],fa[i]=r,i=j;
return r;
}
void combine(int x,int y)
{
int xx=find(x);
int yy=find(y);
if(xx!=yy)
fa[xx]=yy;
}
int main()
{
int k,n,m,i,j,x,y,z;
memset(ma,0,sizeof ma);
scanf("%d%d%d",&k,&n,&m);
for(i=0;i<m;i++)
{
scanf("%d%d%d",&x,&y,&z);
ma[x][y]+=z;
}
for(i=1;i<=n;i++)
{
int num=0,sum=0;
for(j=1;j<=n;j++)
{
if(i==j)continue;
if(ma[i][j]>0&&ma[i][j]<=5)
{
num++;
if(ma[j][i]>0)
sum++;
}
}
//cout<<i<<" num: "<<num<<" sum: "<<sum<<endl;
if(num>k&&sum*5<=num)
sus.push_back(i);
}
//for(i=0;i<sus.size();i++)
//cout<<sus[i]<<endl;
for(i=1;i<=n;i++)
fa[i]=i;
for(i=0;i<sus.size();i++)
{
for(j=0;j<sus.size();j++)
{
if(i==j)continue;
if(ma[sus[i]][sus[j]]&&ma[sus[j]][sus[i]])
{
//cout<<"combine:"<<sus[i]<<" "<<sus[j]<<endl;
combine(sus[i],sus[j]);
}
}
}
vector<int> vec[1005];
bool vis[1005]={0};
for(i=0;i<sus.size();i++)
{
if(vis[sus[i]]==0)
{
vis[sus[i]]=1;
printf("%d",sus[i]);
for(j=0;j<sus.size();j++)
{
if(vis[sus[j]]==0&&find(sus[i])==find(sus[j]))
{
vis[sus[j]]=1;
printf(" %d",sus[j]);
}
}
printf("\n");
}
}
if(sus.size()==0)
printf("None\n");
}