水题一道,bfs
class Solution { public: int vis[110][110]; int dis[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}}; int cal(int x) { int sum = 0; while(x) { sum += x % 10; x /= 10; } return sum; } void bfs(int &cnt, const int &m, const int &n, const int &k) { queue<pair<int, int> > Q; Q.push(pair<int, int> (0, 0)); vis[0][0] = 1; cnt++; while(!Q.empty()) { pair<int, int> u = Q.front(); Q.pop(); for(int i = 0; i < 4; i++) { int vx = u.first + dis[i][0]; int vy = u.second + dis[i][1]; if(vx < 0 || vy < 0 || vx >= m || vy >= n || vis[vx][vy] || cal(vx) + cal(vy) > k) continue; vis[vx][vy] = 1; cnt++; Q.push(pair<int, int> (vx, vy)); } } } int movingCount(int m, int n, int k) { memset(vis, 0, sizeof(vis)); int cnt = 0; bfs(cnt, m, n, k); return cnt; } };
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