1158 Telefraud Detection – PAT甲级真题

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

题目大意：编写一个程序，从大量的通话记录中检测出电信诈骗嫌疑人。如果一个人每天给不同的人打超过K个短电话，但这些人中回电的比例不超过20%，那他可能是嫌疑人。而且，如果两个嫌疑人互相打电话，我们认为他们可能属于同一个团伙。A给B打短电话是指A到B的通话总时长不超过5分钟。给三个正整数：短电话数量的阈值K，不同电话号码的数量N，通话记录数量M。接下来M行给出一天的通话记录，格式为呼叫者caller、接收者receiver、时长duration。输出格式：在每一行输出一个团伙中所有检测到的嫌疑人，按数字升序排列。这些帮派按第一个成员的升序排列。一行中的数字必须使用1个空格分隔，并且行首或行尾不得有多余空格。
分析：p为并查集的父集，sc中存储短通话的人数，rec中存储短通话收到回电的人数，mark用来标记是否输出过，record记录两个人的通话总时长，su中保存嫌疑人的编号。我们先记录下来两人之间的通话总时间，以此来统计短通话以及回电的次数，进而判断出某个人是不是嫌疑人。如果没有嫌疑人，则输出None。因为要按照帮派输出，所以我们用并查集把互相通过电话的嫌疑人设置到同一个集合内，然后就可以输出答案了~

#include <iostream>
#include <vector>
using namespace std;
int k, n, m, c, r, d, p[1001], sc[1001], rec[1001], mark[1001], record[1001][1001];
vector<int> su;
int Find(int a) {
    if (p[a] != a) return p[a] = Find(p[a]);
    return a;
}
void add(int a, int b) {
    int f1 = Find(a), f2 = Find(b);
    if (f1 < f2) p[f2] = f1;
    else p[f1] = f2;
}
int main() {
    for (int i = 1; i <= 1000; i++) p[i] = i;
    cin >> k >> n >> m;
    for (int i = 0; i < m; i++) {
        cin >> c >> r >> d;
        record[c][r] += d;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (record[i][j] && record[i][j] <= 5) {
                sc[i]++;
                if (record[j][i]) rec[i]++;
            }
        }
        if (sc[i] > k && rec[i] * 5 <= sc[i]) su.push_back(i);
    }
    if (su.empty()) {
        cout << "None";
        return 0;
    }
    for (int i = 0; i < su.size(); i++) {
        for (int j = i + 1; j < su.size(); j++) {
            if (record[su[i]][su[j]] && record[su[j]][su[i]]) add(su[i], su[j]);
        }
    }
    for (int i = 0; i < su.size(); i++) {
        if (mark[su[i]]) continue;
        cout << su[i];
        for (int j = i + 1; j < su.size(); j++) {
            if (Find(su[i]) == Find(su[j])) {
                cout << ' ' << su[j];
                mark[su[j]] = 1;
            }
        }
        cout << '\n';
    }
    return 0;
}

#include <iostream>

#include <vector>

using namespace std;

int k, n, m, c, r, d, p[1001], sc[1001], rec[1001], mark[1001], record[1001][1001];

vector<int> su;

int Find(int a) {

if (p[a] != a) return p[a] = Find(p[a]);

return a;

}

void add(int a, int b) {

int f1 = Find(a), f2 = Find(b);

if (f1 < f2) p[f2] = f1;

else p[f1] = f2;

}

int main() {

for (int i = 1; i <= 1000; i++) p[i] = i;

cin >> k >> n >> m;

for (int i = 0; i < m; i++) {

cin >> c >> r >> d;

record[c][r] += d;

}

for (int i = 1; i <= n; i++) {

for (int j = 1; j <= n; j++) {

if (record[i][j] && record[i][j] <= 5) {

sc[i]++;

if (record[j][i]) rec[i]++;

}

if (sc[i] > k && rec[i] * 5 <= sc[i]) su.push_back(i);

}

if (su.empty()) {

cout << "None";

return 0;

}

for (int i = 0; i < su.size(); i++) {

for (int j = i + 1; j < su.size(); j++) {

if (record[su[i]][su[j]] && record[su[j]][su[i]]) add(su[i], su[j]);

}

for (int i = 0; i < su.size(); i++) {

if (mark[su[i]]) continue;

cout << su[i];

for (int j = i + 1; j < su.size(); j++) {

if (Find(su[i]) == Find(su[j])) {

cout << ' ' << su[j];

mark[su[j]] = 1;

}

cout << '\n';

}

return 0;

}

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