题目传送门
1 /*
2 记忆化搜索(DFS+DP)：dp[x][y] 表示x个蛋，在y楼扔后所需要的实验次数
3 ans = min (ans, max (dp[x][y-i], dp[x-1][i-1]) + 1)；前者表示蛋没碎，则往高处(y-i)搜索
4 后者表示蛋碎了，往低处(i-1)方向搜索
5 这样写不好，每次memset (dp)就会超时:(
6 详细解释：http://blog.csdn.net/fulongxu/article/details/27110435
7 */
8 #include <cstdio>
9 #include <iostream>
10 #include <algorithm>
11 #include <cstring>
12 #include <cmath>
13 using namespace std;
14
15 const int MAXN = 1e3 + 10;
16 const int INF = 0x3f3f3f3f;
17 int dp[11][MAXN];
18
19 int DFS(int x, int y)
20 {
21 if (dp[x][y]) return dp[x][y];
22 if (x == 1) {dp[x][y] = y; return y;}
23 if (y <= 2) {dp[x][y] = y; return y;}
24
25 int ans = INF;
26 for (int i=2; i<y; ++i)
27 {
28 int tmp = max (DFS (x, y-i) + 1, DFS (x-1, i-1) + 1);
29 ans = min (ans, tmp);
30 }
31
32 return dp[x][y] = ans;
33 }
34
35 int main(void) //URAL 1223 Chernobyl’ Eagle on a Roof
36 {
37 //freopen ("Z.in", "r", stdin);
38
39 int n, m;
40 while (scanf ("%d%d", &n, &m) == 2)
41 {
42 if (!n && !m) break;
43 if (n > 10) n = 10;
44
45 printf ("%d
", DFS (n, m));
46 }
47
48 return 0;
49 }