[모의 SW 역량테스트] 탈주범 검거

#include<iostream>
#include<queue>
#include<utility>
#include<cstring>
 
using namespace std;
 
int tc,n,m,R,C,L;
int map[51][51];
int d[51][51];
bool dist[51][51];
int dx[8][4] = {{0,0,0,0}, {0,0,1,-1},{1,-1,0,0},{0,0,0,0},{-1,0,0,0},{1,0,0,0},
{1,0,0,0},{-1,0,0,0 }};
int dy[8][4] = { {0,0,0,0}, {1,-1,0,0},{0,0,0,0},{1,-1,0,0},{0,1,0,0},{0,1,0,0},
{0,-1,0,0},{0,-1,0,0}};
int ans;
 
bool canbe(int nx, int ny,int x, int y) {
    for (int i = 0; i < 4; i++) {
        if (map[nx][ny] == 0) continue;
        int sx = nx + dx[map[nx][ny]][i]; int sy = ny + dy[map[nx][ny]][i];
        if (sx >= 0 && sx < n && sy >= 0 && sy < m) {
            if (sx == x && sy == y) {
                return true;
            }
        }
 
    }
    return false;
}
 
void bfs(int x, int y) {
    queue<pair<int, int> > q;
    q.push(make_pair(x, y));
    int s = 1;
    dist[x][y] = true;
    d[x][y] = 1;
    if (L > 1) {
        while (!q.empty()) {
            x = q.front().first; y = q.front().second;
            int val = map[x][y];
            q.pop();
            for (int i = 0; i < 4; i++) {
                int nx = x + dx[val][i]; int ny = y + dy[val][i];
                if (nx >= 0 && nx < n && ny >= 0 && ny < m && map[nx][ny] != 0) {
                    if (d[nx][ny] == 0 && canbe(nx, ny, x, y) && dist[nx][ny] == false) {
                        d[nx][ny] = d[x][y] + 1;
                        dist[nx][ny] = true;
                        s++;
                        if (d[nx][ny] < L) {
                            q.push(make_pair(nx, ny));
                        }
                    }
                }
            }
 
        }
    }
    ans = s;
}
 
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> tc;
    for (int t = 1; t <= tc; t++) {
        cin >> n >> m >> R >> C >> L;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                cin >> map[i][j];
            }
        }
        bfs(R, C);
 
        cout << '#' <<t<< ' ' << ans << '\n';
        memset(d, 0, sizeof(d));
        memset(dist, false, sizeof(dist));
        memset(map, 0, sizeof(map));
    }
    return 0;
}

알고리즘에서 BFS,DFS 개념 자체는 어렵지않다.

어려운것은 조건을 잘보고 그에 따라 구현할줄알아야한다

이번에는 시간 L 이 1일때를 고려하지않아서 한번 틀린문제