解法一

广度优先搜索，使用pair，以及队列queue，先进先出

//广度优先搜索
class Solution {
    int get(int x){
        int res = 0;
        for ( ; x; x/=10)
        {
            res+=x % 10;
        }
        return res;
    }
public:
    int movingCount(int m, int n, int k) {
        if(!k) return 1;
        queue<pair<int,int>> Q; //先进先出
        //向左和向下数组
        int dx[2] = {1,0};
        int dy[2] = {0,1};
        vector<vector<int>> vis(m,vector<int>(n,0));
        Q.push(make_pair(0,0));
        vis[0][0] = 1;
        int ans =1 ;
        while(!Q.empty()){
            auto [x,y] = Q.front();
            Q.pop();
            for (int i = 0; i < 2; i++)
            {
                int tx = dx[i]+x;
                int ty = dy[i]+y;
                if(tx<0 || tx>=m || ty<0 || ty>=n|| vis[tx][ty]|| get(tx)+get(ty)>k) continue;
                Q.push(make_pair(tx,ty));
                vis[tx][ty] = 1;
                ans++;
            }
        }
        return ans;
    }
};

解法二

递推，找到上一个可达的格子，推断下一个是否可达，然后叠加ans

class Solution {
    int get(int x){
        int res = 0;
        for ( ; x; x = x/10)
        {
            res += x%10;
        }
        return res;
    }
 
public:
    int movingCount(int m, int n, int k) {
        if(!k) return 1;
        int ans = 1;
        vector<vector<int>> vis(m,vector<int>(n,0));
        vis[0][0]=1;
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if((i == 0 && j ==0) || get(i)+get(j)>k) continue;
                if(i-1>=0) vis[i][j] = vis[i][j] || vis[i-1][j];
                if(j-1>=0) vis[i][j] = vis[i][j] || vis[i][j-1];
                ans+=vis[i][j];
            }
        }
        return ans;
    }
};