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二蛋问题：

给你2个鸡蛋。
您可以进入一栋 100 层高的建筑。
鸡蛋可能非常坚硬或非常脆弱，这意味着如果从一楼掉下它可能会破裂，或者如果从 100 楼掉下来甚至可能不会破裂。两个鸡蛋是相同的。
你需要弄清楚一座 100 层高的建筑物的最高楼层，一个鸡蛋可以掉下来而不会破裂。
现在的问题是你需要做多少滴。在此过程中，您可以打碎 2 个鸡蛋。

我知道动态编程解决这个问题的方法。我想跟踪解决方案以及最少的尝试次数。即我必须尝试获得最少尝试次数的楼层。

# 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 worst
  case with n eggs and k floors */
int eggDrop(int n, int k)
{
    /* A 2D table where entery eggFloor[i][j] will represent minimum
       number 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 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] = 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 result
    return 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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1 回答 1

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您只需要存储为您提供最佳解决方案的 x 值：

int eggDrop(int n, int k)
{
    /* A 2D table where entery eggFloor[i][j] will represent minimum
       number of trials needed for i eggs and j floors. */
    int eggFloor[n+1][k+1];
    int floor[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] = 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;
                    floor[i][j] = x;
                }                        
            }
        }
    }

    // eggFloor[n][k] holds the result
    return eggFloor[n][k];
}

最后，包含有鸡蛋和地板floor[i][j]时需要尝试的地板。ij

于 2013-10-27T08:38:56.543 回答