剑指13：机器人的运动范围

传送门

nowcoder
leetcode

题目描述

地上有一个 m 行 n 列的方格。一个机器人从坐标 (0, 0) 的格子开始移动，
每一次只能向左，右，上，下四个方向移动一格，但是不能进入行坐标和列坐标的数位之和大于 k 的格子。
例如，当 k 为 18 时，机器人能够进入方格 (35, 37)，因为 3+5+3+7 = 18 。
但是，它不能进入方格 (35, 38)，因为 3+5+3+8 = 19 。
请问该机器人能够达到多少个格子？

C++ 代码 - nowcoder

/*
    递归
*/
class Solution {
public:
    int movingCount(int threshold, int rows, int cols)
    {
        if (rows < 0 || cols < 0) return 0;

        vector<vector<bool>> visited(rows, vector<bool>(cols, false));
        return movingCountCore(threshold, rows, cols, visited, 0, 0);
    }

    int movingCountCore(int threshold, int rows, int cols, 
        vector<vector<bool>> &visited, int i, int j)
    {
        if (i < 0 || i >= rows || j < 0 || j >= cols 
            || !canReach(threshold, i, j) || visited[i][j] == true)
            return 0;
        
        visited[i][j] = true;
        // 上右下左
        return movingCountCore(threshold, rows, cols, visited, i - 1, j) 
            + movingCountCore(threshold, rows, cols, visited, i, j + 1) 
            + movingCountCore(threshold, rows, cols, visited, i + 1, j) 
            + movingCountCore(threshold, rows, cols, visited, i, j - 1) 
            + 1;
    }

    bool canReach(int threshold, int x, int y) {
        int sum = 0;
        while (x > 0) {
            sum += x % 10;
            x /= 10;
        }
        while (y > 0) {
            sum += y % 10;
            y /= 10;
        }
        return sum <= threshold;
    }
};

/*
    BFS
*/
class Solution {
public:
    int movingCount(int threshold, int rows, int cols)
    {
        if (rows < 0 || cols < 0) return 0;

        vector<vector<bool>> visited(rows, vector<bool>(cols, false));
        queue<pair<int, int>> q;
        q.push(make_pair(0, 0));
        visited[0][0] = true;

        int dx[] = { -1, 0, 1, 0 }, dy[] = { 0, 1, 0, -1 }; // 上右下左
        int cnt = 0;
        while (!q.empty()) {
            cnt ++;
            // int x, y;
            // tie(x, y) = q.front();
            int x = q.front().first, y = q.front().second;
            q.pop();
            for (int i = 0; i < 4; i ++) {
                int a = x + dx[i], b = y + dy[i];
                if (a >= 0 && a < rows && b >= 0 && b < cols 
                    && !visited[a][b] && canReach(threshold, a, b)) {
                    q.push(make_pair(a, b));
                    visited[a][b] = true;
                }
            }
        }
        return cnt;
    }

    bool canReach(int threshold, int x, int y) {
        int sum = 0;
        while (x > 0) {
            sum += x % 10;
            x /= 10;
        }
        while (y > 0) {
            sum += y % 10;
            y /= 10;
        }
        return sum <= threshold;
    }
};

C++ 代码 - leetcode

/*
    递归
*/
class Solution {
public:
    int movingCount(int m, int n, int k) {
        if (m < 0 || n < 0) return 0;

        vector<vector<bool>> visited(m, vector<bool>(n, false));
        return dfs(m, n, k, visited, 0, 0);
    }

    int dfs(int rows, int cols, int threshold, 
        vector<vector<bool>> &visited, int i, int j) {
        if (i < 0 || i >= rows || j < 0 || j >= cols || !canReach(i, j, threshold) || visited[i][j]) {
            return 0;
        } 

        visited[i][j] = true;
        // 上右下左
        return dfs(rows, cols, threshold, visited, i - 1, j) 
            + dfs(rows, cols, threshold, visited, i, j + 1) 
            + dfs(rows, cols, threshold, visited, i + 1, j)
            + dfs(rows, cols, threshold, visited, i, j - 1) + 1;
    }

    bool canReach(int x, int y, int threshold) {
        int sum = 0;
        while (x > 0) {
            sum += x % 10;
            x /= 10;
        }
        while (y > 0) {
            sum += y % 10;
            y /= 10;
        }
        return sum <= threshold;
    }
};

/*
    BFS
*/
class Solution {
public:
    int movingCount(int m, int n, int k) {
        if (m < 0 || n < 0) return 0;

        vector<vector<bool>> visited(m, vector<bool>(n, false));
        queue<pair<int, int>> q;
        q.push(make_pair(0, 0));
        visited[0][0] = true;

        int dx[] = { -1, 0, 1, 0 }, dy[] = { 0, 1, 0, -1 }; // 上右下左
        int cnt = 0;
        while (!q.empty()) {
            // int x, y;
            // tie(x, y) = q.front();
            int x = q.front().first, y = q.front().second;
            q.pop();
            cnt ++;
            for (int i = 0; i < 4; i ++) {
                int a = x + dx[i], b = y + dy[i];
                if (a >= 0 && a < m && b >= 0 && b < n 
                    && !visited[a][b] && canReach(a, b, k)) {
                    q.push(make_pair(a, b));
                    visited[a][b] = true;
                }
            }
        }
        return cnt;
    }

    bool canReach(int x, int y, int threshold) {
        int sum = 0;
        while (x > 0) {
            sum += x % 10;
            x /= 10;
        }
        while (y > 0) {
            sum += y % 10;
            y /= 10;
        }
        return sum <= threshold;
    }
};