[Algorithm] Programmers :: 미로 탈출

https://school.programmers.co.kr/learn/courses/30/lessons/159993

프로그래머스

struct Node
{
    int y;
    int x;
    int count;
};
 
bool visited[101][101] = { false };
int dir[4][2] = { {-1,0}, {1,0}, {0,-1}, {0,1} };
 
int Find(int startY, int startX, char target, vector<string>& maps)
{
    memset(visited, false, sizeof(visited));
    queue<Node> q;
    q.push({ startY,startX,0 });
 
    while (!q.empty())
    {
        int curY = q.front().y;
        int curX = q.front().x;
        int curCount = q.front().count;
        q.pop();
 
        if (maps[curY][curX] == target)
        {
            return curCount;
        }
 
        for (int i = 0; i < 4; ++i)
        {
            int nextY = curY + dir[i][0];
            int nextX = curX + dir[i][1];
 
            if (nextY < 0 || nextY >= maps.size()) continue;
            if (nextX < 0 || nextX >= maps[0].size()) continue;
            if (maps[nextY][nextX] == 'X') continue;
            if (visited[nextY][nextX]) continue;
 
            visited[nextY][nextX] = true;
            q.push({ nextY,nextX,curCount + 1 });
        }
    }
    
 
    return -1;
}
 
int solution(vector<string> maps) {
    int answer = 0;
 
    Node start, lever;
 
    for (int i = 0; i < maps.size(); ++i)
    {
        for (int j = 0; j < maps[i].size(); ++j)
        {
            if (maps[i][j] == 'S') start = { i,j };
            else if (maps[i][j] == 'L') lever = { i,j };
        }
    }
 
    int toLever = Find(start.y, start.x, 'L',maps);
    int toEscape = Find(lever.y, lever.x, 'E', maps);
    
    if (toLever == -1 || toEscape == -1) return -1;
 
    return toLever + toEscape;
}

BFS 기본예제로 적합한 것 같다.