PAT 2019-3 7-3 Telefraud Detection

Description:

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. 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（閾值） 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:

• 模擬題

Attention:

• A1034的升級版

Code:

/*
嫌疑人：與超過K個不同的人撥打短電話（通話總時長<=5），但是<=20%的人回復電話
犯罪團伙：兩個嫌疑人之間互通電話

閾值K<=500，通話人數N<=1e3，通話記錄數量M<=1e5
打電話者（1~N），接電話者，通話時長

按行打印犯罪團伙（第一個嫌疑人的序號遞增
嫌疑人按照序號遞增，

基本思路：
圖存儲，通話總時長
二重循環遍歷圖一次，記錄每個人往外撥打短電話的人數，並統計其中回復的人數
如果，超過K個人，並且回復的人數re<=0.2*call -> 5*re <= call，認定為嫌疑人
存儲至嫌疑人隊列

二重循環遍歷嫌疑人隊列，判斷任意兩個人是否有過通話
如果有，則Union，並領較小者為父親結點

再次遍歷嫌疑人隊列，存儲頭目，以頭目為下標，存儲團伙成員

打印，遍歷頭目，輸出團伙成員
*/
#include<cstdio>
#include<vector>
#include<set>
using namespace std;
const int M=1e3+10;
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[0],grap[0]+M*M,0);
    for(int i=0; i<M; i++)
        father[i]=i;

    int k,n,m;
    scanf("%d%d%d", &k,&n,&m);
    for(int i=0; i<m; i++)
    {
        int v1,v2,w;
        scanf("%d%d%d", &v1,&v2,&w);
        grap[v1][v2] += w;
    }
    vector<int> suspect;
    for(int i=1; i<=n; i++)
    {
        int call=0,re=0;
        for(int j=1; j<=n; j++)
        {
            if(grap[i][j]!=0 && grap[i][j]<=5)
            {
                call++;
                if(grap[j][i]!=0)
                    re++;
            }
        }
        if(call>k && 5*re<=call)
            suspect.push_back(i);
    }
    for(int i=0; i<suspect.size(); i++)
    {
        for(int j=i+1; j<suspect.size(); j++)
        {
            int s1=suspect[i],s2=suspect[j];
            if(grap[s1][s2]!=0 && grap[s2][s1]!=0)
                Union(s1,s2);
        }
    }
    set<int> head;
    for(int i=0; i<suspect.size(); i++)
    {
        int s = Find(suspect[i]);
        head.insert(s);
        gang[s].push_back(suspect[i]);
    }
    if(head.size()==0)
        printf("None");
    else
    {
        for(auto it=head.begin(); it!=head.end(); it++)
            for(int i=0; i<gang[*it].size(); i++)
                printf("%d%c", gang[*it][i],i==gang[*it].size()-1?'\n':' ');
    }

    return 0;
}