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为了解决这个问题，我们需要判断狗狗是否能够在指定的时间T到达迷宫中的门。迷宫中的每个位置可以用0或1表示，奇偶性不同的位置相邻。狗狗只能在特定时间T到达门，且每秒只能移动一次，不能重复经过任何一个格子。

方法思路

解决代码

#include 
   
    
#include 
    
     
using namespace std;
int n, m, t;
char maze[11][11]; // 1-based indexing
int startx, starty, endx, endy;
bool flag = false;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
void dfs(int x, int y, int step) {
    if (flag) return;
    if (x == endx && y == endy && step == t) {
        flag = true;
        return;
    }
    if (step > t) return;
    for (int k = 0; k < 4; ++k) {
        int nx = x + dx[k];
        int ny = y + dy[k];
        if (nx < 1 || nx > n || ny < 1 || ny > m) continue;
        if (maze[nx][ny] == 'X') continue;
        int required_step = step + 1;
        int start_odd = (startx + starty) % 2;
        int end_odd = (endx + endy) % 2;
        if ((start_odd == end_odd && (required_step % 2 != 0)) ||
            (start_odd != end_odd && (required_step % 2 == 0))) {
            continue;
        }
        if (visited[nx][ny][required_step]) continue;
        visited[nx][ny][required_step] = true;
        dfs(nx, ny, required_step);
    }
}
int main() {
    while (true) {
        cin >> n >> m >> t;
        if (n == 0 && m == 0 && t == 0) break;
        for (int i = 1; i <= n; ++i) {
            string s;
            cin >> s;
            for (int j = 1; j <= m; ++j) {
                maze[i][j] = s[j-1];
                if (maze[i][j] == 'S') {
                    startx = i;
                    starty = j;
                } else if (maze[i][j] == 'D') {
                    endx = i;
                    endy = j;
                }
            }
        }
        int start_odd = (startx + starty) % 2;
        int end_odd = (endx + endy) % 2;
        bool found = false;
        bool visited[n+1][m+1][t+2];
        for (int x = 1; x <= n; ++x) {
            for (int y = 1; y <= m; ++y) {
                for (int step = 0; step <= t; ++step) {
                    visited[x][y][step] = false;
                }
            }
        }
        queue
     
      
       > q;
        q.push({startx, starty, 0});
        visited[startx][starty][0] = true;
        while (!q.empty()) {
            auto current = q.front();
            q.pop();
            int x = current.first, y = current.second, step = current.third;
            if (x == endx && y == endy && step == t) {
                found = true;
                break;
            }
            for (int k = 0; k < 4; ++k) {
                int nx = x + dx[k];
                int ny = y + dy[k];
                if (nx < 1 || nx > n || ny < 1 || ny > m) continue;
                if (maze[nx][ny] == 'X') continue;
                int required_step = step + 1;
                if ((start_odd == end_odd && (required_step % 2 != 0)) ||
                    (start_odd != end_odd && (required_step % 2 == 0))) {
                    continue;
                }
                if (visited[nx][ny][required_step]) continue;
                visited[nx][ny][required_step] = true;
                q.push({nx, ny, required_step});
            }
        }
        if (found) {
            cout << "YES";
        } else {
            cout << "NO";
        }
    }
}
      
     
    
   

代码解释

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