剑指offer13.机器人的运动范围（中等）

一开始的误解：
以为满足数位之和<=k的地方都能到，实际上有可能是到不了的，因为不是x+y而是x的每一位和y的每一位相加！！！
思路（就是二维数组中地图中相邻1的个数，套模板即可，模板记不清了）
1：BFS

int xy[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
bool flag[m][n];
queue<pair<int, int>> q;
q.push(make_pair(0, 0));
flag[0][0] = true;
while (!q.empty()) {
	pair<int, int> top = q.front();
	//对队头元素操作
	q.pop();
	for (int i = 0; i < 4; ++i) {
		int xx = top.first + xy[i][0];
		int yy = top.second + xy[i][1];
		if (xx < 0 || xx > m - 1 || yy < 0 || yy >n - 1) continue;
		if (map[xx][yy]满足条件 && !flag[xx][yy]) {
			q.push(make_pair(xx, yy));
			flag[xx][yy] = true;
		}
	} 
}

2：DFS

void dfs(int x, int y) {
	if (map[x][y]满足条件 && !flag[x][y]) { //先对当前坐标进行处理！！！
		flag[xx][yy] = true;
	}
	for (int i = 0; i < 4; ++i) {
		int xx = x + xy[i][0];
		int yy = y + xy[i][1];
		if (xx < 0 || xx > m - 1 || yy < 0 || yy >n - 1) continue;
		if (map[xx][yy]满足条件 && !flag[xx][yy]) {
			dfs(xx, yy);
			flag[xx][yy] = true;
		}
	}
}

题目代码粘贴（BFS）

class Solution {
public:
    int xy[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
    int getsum(int m){
        int sum=0;
        while(m){
            sum += (m%10);
            m /= 10;
        }
        return sum;
    }
    int movingCount(int m, int n, int k) {

        vector<vector<bool>> v(m, vector<bool>(n)); //能移位标记为true
        vector<vector<bool>> flag(m, vector<bool>(n)); //搜索过标记为true
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int sum1 = getsum(i), sum2 = getsum(j);
                if (sum1 + sum2 <= k) {
                    v[i][j] = true;
                }
            }
        }
        int ans = 0;
        queue<pair<int, int>> q;
        q.push(make_pair(0, 0));
        flag[0][0] = true;
        while (!q.empty()) {
            pair<int, int> top = q.front();
            ans++;
            q.pop();
            for (int i = 0; i < 4; ++i) {
                int xx = top.first + xy[i][0];
                int yy = top.second + xy[i][1];
                if (xx < 0 || xx > m - 1 || yy < 0 || yy >n - 1) continue;
                if (v[xx][yy] && !flag[xx][yy]) {
                    q.push(make_pair(xx, yy));
                }
                flag[xx][yy] = true;
            } 
        }
        return ans;
    }
};

题目代码粘贴（DFS）

class Solution {
public:
    int xy[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
    int v[105][105];
    int flag[105][105];

    int getsum(int m){
        int sum=0;
        while(m){
            sum += (m%10);
            m /= 10;
        }
        return sum;
    }
    void dfs(int x, int y, int &ans, int &m, int &n) {
        if (v[x][y] && !flag[x][y]) {
            ans++;
            flag[x][y] = true;
        }
        for (int i = 0; i < 4; ++i) {
            int xx = x + xy[i][0];
            int yy = y + xy[i][1];
            if (xx < 0 || xx > m - 1 || yy < 0 || yy > n - 1) continue;
            if (v[xx][yy] && !flag[xx][yy]) {
                ans++;
                flag[xx][yy] = true;
                dfs(xx, yy, ans, m, n);
            }
        }

    }
    int movingCount(int m, int n, int k) {

        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int sum1 = getsum(i), sum2 = getsum(j);
                if (sum1 + sum2 <= k) {
                    v[i][j] = true;
                }
            }
        }

        int ans = 0;
        dfs(0, 0, ans, m, n);
        return ans;
    }
};

标签：int,机器人,中等,flag,yy,offer13,xx,xy,true
来源： https://blog.csdn.net/zhangjiaji111/article/details/120710116