프로그래머스 2020 kakao 자물쇠와 열쇠 문제 풀이

#include <string>
#include <vector>

using namespace std;
void makeMyKey(vector<vector<int>>& key);
void makeMyTempKey(vector<vector<int>>& tempKey, int dir, int x, int y, int keysize, int locksize);
bool check(vector<vector<int>>& tempKey, vector<vector<int>>& lock);
int mykey[4][32][32];
bool solution(vector<vector<int>> key, vector<vector<int>> lock) {
	makeMyKey(key);
	vector<vector<int>> tempKey(lock.size(), vector<int>(lock.size(), 0));
	for (int i = 0; i <= key.size() + lock.size(); i++) {
		for (int j = 0; j <= key.size() + lock.size(); j++) {
			for (int k = 0; k < 4; k++) {
				makeMyTempKey(tempKey, k, j, i, key.size(), lock.size());
				if (check(tempKey,lock)) {
					return true;
				}
			}
		}
	}
	return false;
}
void makeMyKey(vector<vector<int>>& key)
{
	int length = key.size();
	for (int i = 0, _i = 0; i < length; i++, _i++)
		for (int j = 0, _j = 0; j < length; j++, _j++)
			mykey[0][i][j] = key[i][j];
	for (int i = 0, _i = 0; i < length; i++, _i++)
		for (int j = 0, _j = length - 1; j < length; j++, _j--)
			mykey[1][i][j] = key[_j][_i];
	for (int i = 0, _i = 0; i < length; i++, _i++)
		for (int j = 0, _j = length - 1; j < length; j++, _j--)
			mykey[2][i][j] = mykey[1][_j][_i];
	for (int i = 0, _i = 0; i < length; i++, _i++)
		for (int j = 0, _j = length - 1; j < length; j++, _j--)
			mykey[3][i][j] = mykey[2][_j][_i];
}
void makeMyTempKey(vector<vector<int>>& tempKey, int dir, int x, int y, int keysize, int locksize) {
	int keyX = keysize - x;
	int keyY = keysize - y;

	for (int i = 0,_i = keyY; i < locksize; i++, _i++) {
		for (int j = 0, _j = keyX; j < locksize; j++, _j++)
		{
			if (_j >= keysize || _j < 0 || _i >= keysize || _i < 0) {
				tempKey[i][j] = 0;
			}
			else {
				tempKey[i][j] = mykey[dir][_i][_j];
			}
		}

	}
}
bool check(vector<vector<int>>& tempKey, vector<vector<int>>& lock) {
	int length = lock.size();
	for (int i = 0; i < length; i++)
	{
		for (int j = 0; j < length; j++)
		{
			if (lock[i][j] + tempKey[i][j] == 0 || lock[i][j] + tempKey[i][j] == 2)
				return false;
		}
	}
	return true;
}