C++:: 프로그래머스 < 당구 연습 >

개인적으로 생각하기 좀 어렵고 번거로웠던 문제이다. 치는공과 맞춰야하는 공의 위치를 잘 선정해서 풀면 쉽게 풀수 있다. 필자의 풀이는 별로이니 아래 코드를 참고하는것을 추천한다...

#include <string>
#include <vector>

using namespace std;

int pow(int n)
{
    return n*n;
}

vector<int> solution(int m, int n, int startX, int startY, vector<vector<int>> balls)
{
    vector<int> answer;
    int cnt = balls.size();
    for(int i=0;i<cnt;i++)
    {
        int len, a, b;
        if (balls[i][0] == startX)
        {
            a = abs(startY - balls[i][1]);
            b = 2 * min(m - startX, startX);
            if (startY > balls[i][1]) len = pow(a + 2*(n - startY));
            else len = pow(a + 2*startY);
        }
        else if(balls[i][1] == startY)
        {
            a = abs(startX - balls[i][0]);
            b = 2 * min(n - startY, startY);
            if (startX > balls[i][0]) len = pow(a + 2*(m - startX));
            else len = pow(a + 2*startX);
        }
        else
        {
            a = min(startX + balls[i][0], 2*m - startX - balls[i][0]);
            b = abs(startY - balls[i][1]);
            len = pow(a) + pow(b);
            a = min(startY + balls[i][1], 2*n - startY - balls[i][1]);
            b = abs(startX - balls[i][0]);
        }
        len = min(len, pow(a) + pow(b));
        answer.push_back(len);
    }
    return answer;
}

좋아보였던 풀이!!!

#include <cmath>
#include <vector>
#include <climits>

using namespace std;

int calc_dist(int m, int n, int a, int b, int c, int d)
{
    int total=INT_MAX;
    if(a!=c || b<=d) total=min(total, (int)(pow(a-c, 2)+pow(b+d, 2)));
    if(a>=c || b!=d) total=min(total, (int)(pow(a-2*m+c, 2)+pow(b-d, 2)));
    if(a!=c || b>=d) total=min(total, (int)(pow(a-c, 2)+pow(b-2*n+d, 2)));
    if(a<=c || b!=d) total=min(total, (int)(pow(a+c, 2)+pow(b-d, 2)));
    return total;
}

vector<int> solution(int m, int n, int startX, int startY, vector<vector<int>> balls)
{
    vector<int> answer;

    for(int i=0; i<balls.size(); i++)
        answer.push_back(calc_dist(m, n, startX, startY, balls[i][0], balls[i][1]));

    return answer;
}