PAT 2019-3 7-3 Telefraud Detection

Description:

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. Amakes 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 (≤, the threshold（阈值） of the amount of short phone calls), N (≤, the number of different phone numbers), and M (≤, 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

Keys:

• 模拟题

Attention:

• A1034的升级版

Code:

 /*
 嫌疑人：与超过K个不同的人拨打短电话（通话总时长<=5），但是<=20%的人回复电话
 犯罪团伙：两个嫌疑人之间互通电话

 阈值K<=500，通话人数N<=1e3，通话记录数量M<=1e5
 打电话者（1~N），接电话者，通话时长

 按行打印犯罪团伙（第一个嫌疑人的序号递增
 嫌疑人按照序号递增，

 基本思路：
 图存储，通话总时长
 二重循环遍历图一次，记录每个人往外拨打短电话的人数，并统计其中回复的人数
 如果，超过K个人，并且回复的人数re<=0.2*call -> 5*re <= call，认定为嫌疑人
 存储至嫌疑人队列

 二重循环遍历嫌疑人队列，判断任意两个人是否有过通话
 如果有，则Union，并领较小者为父亲结点

 再次遍历嫌疑人队列，存储头目，以头目为下标，存储团伙成员

 打印，遍历头目，输出团伙成员
 */
 #include<cstdio>
 #include<vector>
 #include<set>
 using namespace std;
 const int M=1e3+;
 int grap[M][M],father[M];
 vector<int> gang[M];

 int Find(int v)
 {
     int s=v;
     while(v != father[v])
         v=father[v];
     while(s != father[s])
     {
         int x = father[s];
         father[s] = v;
         s = x;
     }
     return v;
 }

 void Union(int s1, int s2)
 {
     int f1=Find(s1);
     int f2=Find(s2);
     if(f1<f2)
         father[f2]=f1;
     else
         father[f1]=f2;
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDEG

     fill(grap[],grap[]+M*M,);
     for(int i=; i<M; i++)
         father[i]=i;

     int k,n,m;
     scanf("%d%d%d", &k,&n,&m);
     for(int i=; i<m; i++)
     {
         int v1,v2,w;
         scanf("%d%d%d", &v1,&v2,&w);
         grap[v1][v2] += w;
     }
     vector<int> suspect;
     for(int i=; i<=n; i++)
     {
         int call=,re=;
         for(int j=; j<=n; j++)
         {
             if(grap[i][j]!= && grap[i][j]<=)
             {
                 call++;
                 if(grap[j][i]!=)
                     re++;
             }
         }
         if(call>k && *re<=call)
             suspect.push_back(i);
     }
     for(int i=; i<suspect.size(); i++)
     {
         for(int j=i+; j<suspect.size(); j++)
         {
             int s1=suspect[i],s2=suspect[j];
             if(grap[s1][s2]!= && grap[s2][s1]!=)
                 Union(s1,s2);
         }
     }
     set<int> head;
     for(int i=; i<suspect.size(); i++)
     {
         int s = Find(suspect[i]);
         head.insert(s);
         gang[s].push_back(suspect[i]);
     }
     if(head.size()==)
         printf("None");
     else
     {
         for(auto it=head.begin(); it!=head.end(); it++)
             for(int i=; i<gang[*it].size(); i++)
                 printf("%d%c", gang[*it][i],i==gang[*it].size()-?'\n':' ');
     }

     return ;
 }