记忆化搜索(DFS+DP) URAL 1223 Chernobyl’ Eagle on a Roof

 

题目传送门

/*
    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
        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
            后者表示蛋碎了，往低处(i-1)方向搜索
    这样写不好，每次memset (dp)就会超时:(
    详细解释：http://blog.csdn.net/fulongxu/article/details/27110435        
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

const int MAXN = 1e3 + 10;
const int INF = 0x3f3f3f3f;
int dp[11][MAXN];

int DFS(int x, int y)
{
    if (dp[x][y])    return dp[x][y];
    if (x == 1)    {dp[x][y] = y;    return y;}
    if (y <= 2)    {dp[x][y] = y;    return y;}

    int ans = INF;
    for (int i=2; i<y; ++i)
    {
        int tmp = max (DFS (x, y-i) + 1, DFS (x-1, i-1) + 1);
        ans = min (ans, tmp);
    }

    return dp[x][y] = ans;
}

int main(void)        //URAL 1223 Chernobyl’ Eagle on a Roof
{
    //freopen ("Z.in", "r", stdin);

    int n, m;
    while (scanf ("%d%d", &n, &m) == 2)
    {
        if (!n && !m)    break;
        if (n > 10)    n = 10;

        printf ("%d\n", DFS (n, m));
    }

    return 0;
}