并查集应用——PAT甲级2019春季

2019/7/24 10:07:36 人评论 次浏览 分类:学习教程

并查集适用问题举例

1、已知,有n个人和m对好友关系
2、如果两个人是直接的或者间接的好友(好友的好友的好友。。。),那么他们属于一个集合,就是一个朋友圈中
3、写出程序,求这n个人中一共有多少个朋友圈

更详细请查看

http://www.omegaxyz.com/2019/02/25/union-find/

PAT-2019年春季考试-甲级-3

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 (≤10​3​​, the number of different phone numbers), and M (≤10​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

C++代码:

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<map>
 
using namespace std;
 
int durcnt[1010][1010]; //统计总通话时间
int father[1010]; //并查集father
 
int findFather(int n){
    int a = n;
    while(a != father[a])a = father[a];
    while(n != father[n]){
        int z = n;
        n = father[n];
        father[z] = a;
    }
    return a;
}
 
void setUnion(int x, int y){
    int fx = findFather(x);
    int fy = findFather(y);
    if(fx != fy){
        if(fy < fx) swap(fx, fy); //序号较小的作为leader
        father[fy] = fx;
    }
}
 
void init(int n){
    for(int i = 1; i <= n; i++)father[i] = i;
}
 
int main(){
    map<int, vector<int> > gangs;
    vector<int> suspects;
    int K, N, M;
    cin>>K>>N>>M;
    init(N);
    int c, r, d;
    for(int i = 0; i < M; i++){
        cin>>c>>r>>d;
        durcnt[c][r] += d;
    }
 
    //判断是否是诈骗团伙
    for(int i = 1; i <= N; i++){
        int sumDiff = 0, rcall = 0;
        for(int j = 1; j <= N; j++){
            if(durcnt[i][j] > 0 && durcnt[i][j] <= 5){
                sumDiff++;
                if(durcnt[j][i] > 0)rcall++;
            }
        }
        if(sumDiff > K && 1.0 * rcall / sumDiff <= 0.2)suspects.push_back(i);
    }
    
    //找出每一个团伙的成员
    for(int i = 0; i < suspects.size(); i++){
        for(int j = i; j < suspects.size(); j++){
            if(durcnt[suspects[i]][suspects[j]] > 0 && durcnt[suspects[j]][suspects[i]] > 0){
                setUnion(suspects[i], suspects[j]); //同一个团伙的合并
            }
        }
    }
    for(int i = 0; i < suspects.size(); i++){
        int leader = findFather(suspects[i]);
        gangs[leader].push_back(suspects[i]);
    }
    if(gangs.size() == 0)cout<<"None"<<endl;
    else{
        for(int i = 0; i < gangs.size(); i++){
            sort(gangs[i].begin(), gangs[i].end());
            for(int j = 0; j < gangs[i].size(); j++){
                printf("%d%s", gangs[i][j], j == gangs[i].size() - 1 ? "\n" : " ");
            }
        }
    }
    return 0;
}

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