A1158 Telefraud Detection（25分）PAT甲级 (Advanced Level) Practice（C++）满分题解【并查集】

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, 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

代码如下：

//检测为嫌疑人：
//1、每天打短电话给超过K个不同的人
//2、不超过5min的记录才会算异常，>5min的不算
//3、不超过20%的接受者才会回复
//4、两个嫌疑人互打，可能属于同一个帮派
#include
#include
using namespace std;
const int MAXN=1010;
int G[MAXN][MAXN];//两点之间有无连线，值为两点的通话时长总和
int father[MAXN],visit[MAXN];
vector gang;
int findfather(int a)
{
    return (a==father[a])?a:(father[a]=findfather(father[a]));
}
void u(int x,int y){
    int fatherx=findfather(x);
    int fathery=findfather(y);
    if(fatherx>k>>n>>m;
    fill(G[0],G[0]+MAXN*MAXN,0);
    fill(visit,visit+MAXN,0);
    for(int i=1;i<=n;i++){
        father[i]=i;
    }
    int c,r,d;
    for(int i=1;i<=m;i++)
    {
        cin>>c>>r>>d;
        G[c][r]+=d;
    }
    for(int i=1;i<=n;i++){
        int shortcall_cnt=0,callback_cnt=0;
        for(int j=1;j<=n;j++){
            if(i==j)continue;
            if(G[i][j]>0&&G[i][j]<=5){
                shortcall_cnt++;
                if(G[j][i]==1)callback_cnt++;//有回电的
            }
        }
        //短电话打给超过K个不同的人,不超过20%的回复率
        if(shortcall_cnt>k&&callback_cnt*5<=shortcall_cnt)gang.push_back(i);
    }
    //合并gang
    for(int i=0;i 
运行结果如下：