鸡蛋掉落拼图的Java程序| DP-11

下面是这个著名的谜题实例的描述，它涉及 n=2 个鸡蛋和一个 k=36 层的建筑物。

假设我们想知道 36 层楼的哪些楼层可以安全地掉落鸡蛋，哪些楼层会导致鸡蛋在落地时破裂。我们做几个假设：

......一个在坠落中幸存下来的鸡蛋可以再次使用。
......必须丢弃破碎的鸡蛋。
…..跌倒的效果对所有鸡蛋都是一样的。
…..如果鸡蛋在掉落时会破裂，那么如果从较高的地板上掉落，它就会破裂。
......如果一个鸡蛋在坠落中幸存下来，那么它会在较短的坠落中幸存下来。
……..不排除一楼窗户破蛋，也不排除36楼不破蛋。

如果只有一个鸡蛋可用，并且我们希望确保获得正确的结果，则只能以一种方式进行实验。从一楼窗户掉下鸡蛋；如果它幸存下来，请将其从二楼的窗户扔下。继续向上直到它破裂。在最坏的情况下，这种方法可能需要 36 次粪便。假设有 2 个鸡蛋可用。保证在所有情况下都有效的最少蛋液数量是多少？
问题实际上并不在于找到关键楼层，而只是决定应该从哪个楼层掉下鸡蛋，以便尽量减少试验的总数。

来源：动态编程维基

动态规划解决方案

Java

// A Dynamic Programming based Java Program for the Egg Dropping Puzzle
class EggDrop {
    // A utility function to get maximum of two integers
    static int max(int a, int b) { return (a > b) ? a : b; }
 
    /* Function to get minimum number of trials needed in worst
    case with n eggs and k floors */
    static int eggDrop(int n, int k)
    {
        /* A 2D table where entry eggFloor[i][j] will represent minimum
       number of trials needed for i eggs and j floors. */
        int eggFloor[][] = new int[n + 1][k + 1];
        int res;
        int i, j, x;
 
        // We need one trial for one floor and0 trials for 0 floors
        for (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
        // property
        for (i = 2; i <= n; i++) {
            for (j = 2; j <= k; j++) {
                eggFloor[i][j] = Integer.MAX_VALUE;
                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 result
        return eggFloor[n][k];
    }
 
    /* Driver program to test to printDups*/
    public static void main(String args[])
    {
        int n = 2, k = 10;
        System.out.println("Minimum number of trials in "+
                           "worst case with "
                           + n + "  eggs and " + k + " floors is " + eggDrop(n, k));
    }
}
/*This code is contributed by Rajat Mishra*/

输出：

Minimum number of trials in worst case with 2  eggs and 10 floors is 4

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