蛋滴跟踪表

Question

两个蛋问题：

给你两个鸡蛋。 你可以进入一栋100层楼高的大楼。 鸡蛋可能很硬，也可能非常脆弱，这意味着如果从一楼掉下来，它可能会破裂，如果从第100层掉下来，甚至可能不会破裂，floor.Both鸡蛋是相同的。 你需要找出一座100层楼的最高楼层，一个鸡蛋可以在不破掉的情况下掉下来。 现在的问题是你需要做多少滴。在这个过程中，你可以打碎两个鸡蛋。

我知道用动态规划解决这个问题的方法。我想要跟踪解决方案和最小尝试次数。也就是我必须尝试的楼层，以获得最少的试试次数。

# 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;
}

Answer 1

您只需存储给您最佳解决方案的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，floor[i][j]包含了当您有i鸡蛋和j地板时需要尝试的地板。