Diademata

난 백트랙킹으로 구현 했는데...

한줄로도 되네? 오진다 오져

code>>

bool check(int b_x, int b_y, int p_x, int p_y, int x, int y)

{

if (b_x < 0 || b_x > 7) return false;

if (b_y < 0 || b_y > 7) return false;

if (b_x == p_x && b_y == p_y) return true;

bool result = false;

result = check(b_x + x, b_y + y, p_x, p_y, x, y);

return result;

}

bool bishopAndPawn(std::string bishop, std::string pawn) {

int x[4] = { -1, 1, -1, 1 };

int y[4] = { 1, 1, -1, -1 };

int b_x = bishop[0] - 'a';

int b_y = bishop[1] - '1';

int p_x = pawn[0] - 'a';

int p_y = pawn[1] - '1';

bool result = false;

for (int i = 0; i< 4; i++)

{

result = check(b_x, b_y, p_x, p_y, x[i], y[i]);

if (result)

break;

}

return result;

}

다른 사람 코드>>

bool bishopAndPawn(std::string b, std::string p) {

return abs(p[0] - b[0]) == abs(p[1] - b[1]);

}