标签：nes == caller specific nbsp %s 矩阵 team col
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 (≤103 10^310
3
?? )?? , the number of different phone numbers), and M (≤105 10^510
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
【声明】
由于此题还未上PAT官网题库，故没有测试集，仅仅是通过了样例，若发现错误，感谢留言指正。
Solution：
判断电话诈骗分子，当一个人与>k个不同的人打短时间【<= 5分钟】通话，且这些人中只有不超过20%的人给回了电话，那么这个人就是诈骗犯；
当相互有通话的诈骗犯是认为同一组；
输出按组内升序，组间按第一个组员升序的格式输出。
使用邻接矩阵记录通过时间【累计通话时间】，然后分别记录每个人满足要求的打出的电话人数以及回电话的人数，最后将相互通话的嫌疑犯视为一组。
注重细节哦。
1 #include <iostream> 2 #include <vector> 3 #include <map> 4 using namespace std; 5 int main() 6 { 7 int k, n, m, t; 8 cin >> k >> n >> m; 9 vector<int>call(n + 1, 0), back(n + 1, 0);//短电话打出去的记录和这些人打回进来的记录 10 vector<vector<int>>v(n + 1, vector<int>(n + 1, 0));//电话记录,记住记录的是累计的通话时间 11 while (m--) 12 { 13 int a, b, c; 14 cin >> a >> b >> c; 15 v[a][b] += c;//算累加通话时间的 16 } 17 for (int i =1; i <= n; ++i) 18 { 19 for (int j = 1; j <= n; ++j) 20 { 21 if (v[i][j]>0 && v[i][j] <= 5)//这通话时间算是短电话 22 { 23 ++call[i];//计算i打出去的人数 24 if (v[j][i] > 0)//这是j回给了i的 25 ++back[i];//计算回给i的人数 26 } 27 } 28 } 29 vector<int>suspect, team(n + 1, 0);//统计诈骗犯的人数和初始化他们的团队号 30 for (int i = 1; i <= n; ++i) 31 { 32 if (call[i] > k && call[i] >= back[i]*5)//打出去的人数大于阈值k，回电话的人数不多于打出去的20%,注意 33 { 34 team[i] = i;//用来算同谋的 35 suspect.push_back(i); 36 } 37 } 38 for (int i = 0; i < suspect.size(); ++i)//这里虽然是双重循环，但是诈骗犯的数量级很小的，所以这里应该不会超时间的，当然，超了的话，就用DFS即可 39 for (int j = i + 1; j < suspect.size(); ++j) 40 if (v[suspect[i]][suspect[j]] > 0 && v[suspect[j]][suspect[i]] > 0)//这两个诈骗人相互打过电话 41 team[suspect[j]] = team[suspect[i]]; 42 map<int, vector<int>>res; 43 for (int i = 1; i <= n; ++i) 44 if (team[i] > 0) 45 res[team[i]].push_back(i);//将属于同一伙的诈骗犯放一组 46 if (suspect.size() == 0) 47 printf("None"); 48 else 49 { 50 for (auto ptr = res.begin(); ptr != res.end(); ++ptr) 51 { 52 for (int i = 0; i < ptr->second.size(); ++i)//由于遍历是从小到大的，所以不用排序 53 printf("%s%d", i == 0 ? "" : " ", ptr->second[i]); 54 printf("\n"); 55 } 56 } 57 return 0; 58 }
PAT甲级【2019年3月考题】——A1158 TelefraudDetection【25】
标签：nes == caller specific nbsp %s 矩阵 team col
原文地址：https://www.cnblogs.com/zzw1024/p/11954289.html
