7-3 Telefraud Detection (25分)
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 <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>
using namespace std;

const int N = 1e3 + 10;

int g[N][N], p[N];
int k, n, m;

int find (int x)
{
    if(p[x] != x) return p[x] = find(p[x]);
    return p[x];
}

int main()
{
    cin >> k >> n >> m;
    while(m --)
    {
        int id, son, time;
        cin >> id >> son >> time;
        g[id][son] += time;
    }

    vector <int> h;
    for(int i = 1; i <= n; i ++)
    {
        int in = 0, back = 0;
        for(int j = 1; j <= n; j ++)
        {
            if(g[i][j] && g[i][j] <= 5)
            {
                in ++; 
                if(g[j][i]) back ++;
            }
        }
        if((double) back / in <= 0.2 && in > k) h.push_back(i);
    }

    if(h.empty()) puts("None");
    else{
        for(auto i : h) p[i] = i;
        for(int i = 0; i < h.size(); i ++)
            for(int j = 0; j < i; j ++)
                if(g[h[i]][h[j]] && g[h[j]][h[i]]) p[find(h[i])] = find(h[j]);
        unordered_map <int, vector <int> > tmp;
        for(auto i : h) tmp[find(i)].push_back(i);
        vector <vector <int> > info;
        for(auto i : tmp) info.push_back(i.second);
        sort(info.begin(), info.end());
        for(auto i : info)
        {
            cout << i[0];
            for(int j = 1; j < i.size(); j ++) cout << " " << i[j];
            puts("");
        }
    }
    return 0;
}