[프로그래머스][Level3] 네트워크 c++

https://programmers.co.kr/learn/courses/30/lessons/43162 

프로그래머스

 

풀이 과정

문제를 딱 보자마자 유니온 파인드 알고리즘이 생각났다.  
풀고보니 유니온 파인드의 기본적인 문제이며, 부모 노드의  
중복되지 않은 숫자의 개수는 몇 개인지 세면 되는 단순한 문제였다.

 

풀이 1

#include <string>
#include <vector>
#include <algorithm>

using namespace std;

int parent[201];

int getParent(int x){
    if(parent[x] == x) return x;
    return parent[x] = getParent(parent[x]);
}

void UnionParent(int a, int b){
    a = getParent(a);
    b = getParent(b);

    if(a > b) parent[a] = b;
    else parent[b] = a;
}

int solution(int n, vector<vector<int>> computers) {
    int answer = 0;
    for(int i = 1; i <= n; i++){
        parent[i] = i;
    }
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            if(i != j){
                if(computers[i][j] == 1) UnionParent(i + 1, j + 1);
            }
        }
    }
    vector<int> v;
    for(int i = 1; i <= n; i++){
        v.push_back(getParent(i));
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());

    answer = v.size();
    return answer;
}

 

풀이 2

#include <string>
#include <vector>

using namespace std;

int parent[201];

int getParent(int x){
    if(parent[x] == x) return x;
    return parent[x] = getParent(parent[x]);
}

void UnionParent(int a, int b){
    a = getParent(a);
    b = getParent(b);

    if(a > b) parent[a] = b;
    else parent[b] = a;
}

int solution(int n, vector<vector<int>> computers) {
    int answer = 0;
    for(int i = 1; i <= n; i++){
        parent[i] = i;
    }
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            if(i != j){
                if(computers[i][j] == 1) UnionParent(i + 1, j + 1);
            }
        }
    }
    vector<int> v;
    for(int i = 1; i <= n; i++){
        v.push_back(getParent(i));
    }
    // sort 오름차순 구현
    for(int i = 0; i < v.size(); i++){
        for(int j = 0; j < v.size() - 1; j++){
            if(v[j] > v[j + 1]){
                int temp = v[j];
                v[j] = v[j + 1];
                v[j + 1] = temp;
            }
        }
    }
    // erase, unique 함수 구현
    int num = 1;
    int temp[201];
    temp[0] = v[0];
    for(int i = 1; i < v.size(); i++){
        if(v[i] != temp[num - 1]){
            temp[num++] = v[i];
        }
    }
    answer = num;
    return answer;
}