Egg Drop 扔鸡蛋

假设有 n=2 个完全一样的鸡蛋，有 k=36 层楼。存在一个楼层 m，把鸡蛋从低于m的楼层扔下去的时候，鸡蛋不会摔破；把鸡蛋从高于等于m的楼层扔下去的时候，鸡蛋会摔破。题目要求，找出临界楼层m，并且扔鸡蛋的次数要做到最少。

思路：

最简单的方法是，从第 1 层往上，一层一层的扔鸡蛋，直到鸡蛋摔碎。这样需要扔鸡蛋 m 次。

下面用递归的方法解决该问题。假设鸡蛋从 x 层扔下去。（1）如果鸡蛋摔破，那么还剩 n-1 个鸡蛋，临界楼层在 [1, x] 之间；（2）如果鸡蛋没破，那么还剩 n 个鸡蛋，临界楼层在 [k, x+1] z之间。设 eggDrop(n,k) 是问题的解，那么递归表达式是

eggDrop(n,k) = 1+ min ( max( eggDrop(n-1, x-1), eggDrop(n, k-x) ) for all x in [1,k] )

代码如下：

# include <stdio.h>
# include <limits.h>// A utility function to get maximum of two integers
int max(int a, int b) { return (a > b)? a: b; }/* Function to get minimum number of trails needed in worstcase with n eggs and k floors */
int eggDrop(int n, int k)
{// If there are no floors, then no trials needed. OR if there is// one floor, one trial needed.if (k == 1 || k == 0)return k;// We need k trials for one egg and k floorsif (n == 1)return k;int min = INT_MAX, x, res;// Consider all droppings from 1st floor to kth floor and// return the minimum of these values plus 1.for (x = 1; x <= k; x++){res = max(eggDrop(n-1, x-1), eggDrop(n, k-x));if (res < min)min = res;}return min + 1;
}/* Driver program to test to pront printDups*/
int main()
{int n = 2, k = 10;printf ("\nMinimum number of trials in worst case with %d eggs and ""%d floors is %d \n", n, k, eggDrop(n, k));return 0;
}

动态规划的代码如下：

# include <stdio.h>
# include <limits.h>// A utility function to get maximum of two integers
int max(int a, int b) { return (a > b)? a: b; }/* Function to get minimum number of trails needed in worstcase with n eggs and k floors */
int eggDrop(int n, int k)
{/* A 2D table where entery eggFloor[i][j] will represent minimumnumber of trials needed for i eggs and j floors. */int eggFloor[n+1][k+1];int res;int i, j, x;// We need one trial for one floor and0 trials for 0 floorsfor (i = 1; i <= n; i++){eggFloor[i][1] = 1;eggFloor[i][0] = 0;}// We always need j trials for one egg and j floors.for (j = 1; j <= k; j++)eggFloor[1][j] = j;// Fill rest of the entries in table using optimal substructure// propertyfor (i = 2; i <= n; i++){for (j = 2; j <= k; j++){eggFloor[i][j] = INT_MAX;for (x = 1; x <= j; x++){res = 1 + max(eggFloor[i-1][x-1], eggFloor[i][j-x]);if (res < eggFloor[i][j])eggFloor[i][j] = res;}}}// eggFloor[n][k] holds the resultreturn eggFloor[n][k];
}/* Driver program to test to pront printDups*/
int main()
{int n = 2, k = 36;printf ("\nMinimum number of trials in worst case with %d eggs and ""%d floors is %d \n", n, k, eggDrop(n, k));return 0;
}

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