HDU 1069 Monkey and Banana

Problem Description：

A group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, and at the mean time, provide the monkey with some blocks. If the monkey is clever enough, it shall be able to reach the banana by placing one block on the top another to build a tower and climb up to get its favorite food.

The researchers have n types of blocks, and an unlimited supply of blocks of each type. Each type-i block was a rectangular solid with linear dimensions (xi, yi, zi). A block could be reoriented so that any two of its three dimensions determined the dimensions of the base and the other dimension was the height. 

They want to make sure that the tallest tower possible by stacking blocks can reach the roof. The problem is that, in building a tower, one block could only be placed on top of another block as long as the two base dimensions of the upper block were both strictly smaller than the corresponding base dimensions of the lower block because there has to be some space for the monkey to step on. This meant, for example, that blocks oriented to have equal-sized bases couldn't be stacked. 

Your job is to write a program that determines the height of the tallest tower the monkey can build with a given set of blocks.

 

Input：

The input file will contain one or more test cases. The first line of each test case contains an integer n,
representing the number of different blocks in the following data set. The maximum value for n is 30.
Each of the next n lines contains three integers representing the values xi, yi and zi.
Input is terminated by a value of zero (0) for n.

 

Output：

For each test case, print one line containing the case number (they are numbered sequentially starting from 1) and the height of the tallest possible tower in the format "Case case: maximum height = height".

 

Sample Input：

1

10 20 30

2

6 8 10

5 5 5

7

1 1 1

2 2 2

3 3 3

4 4 4

5 5 5

6 6 6

7 7 7

5

31 41 59

26 53 58

97 93 23

84 62 64

33 83 27

0

 

Sample Output：

Case 1: maximum height = 40

Case 2: maximum height = 21

Case 3: maximum height = 28

Case 4: maximum height = 342

题意：猴子想摘香蕉，现在它有一些长方体，可以搭建在一起帮助它摘到香蕉，现在问这些长方体最高能有多高，搭建的时候有要求，上面的长方体的长和宽必须小于下面长方体的长和宽。

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int INF=0x3f3f3f3f;
const int N=110;

struct node
{
    int a, b, c, s; ///存放长方体的长，宽，高，及底面面积
}no[N];

int cmp(const void *a, const void *b)
{
    node *s1 = (node *)a, *s2 = (node *)b;
    return s2->s - s1->s;
}

int main ()
{
    int n, a, b, c, i, j, k, A[N], ans, num = 0;

    while (scanf("%d", &n), n)
    {
        k = 0;
        ans = -INF;
        num++;

        for (i = 1; i <= n; i++)
        {
            scanf("%d%d%d", &a, &b, &c);

            no[k].a = a;
            no[k].b = b;
            no[k].c = c;
            no[k++].s = a*b;

            no[k].a = a;
            no[k].b = c;
            no[k].c = b;
            no[k++].s = a*c; ///长方体的长，宽，高是不确定的

            no[k].a = b;
            no[k].b = c;
            no[k].c = a;
            no[k++].s = b*c;
        }

        qsort(no, k, sizeof(no[0]), cmp); ///因为上面的长方体长和宽必须大于下面的长方体，所以这些长方体的面积肯定是从大到小排的

        for (i = 0; i < 3*n; i++)
            A[i] = no[i].c; ///可能放一个长方体的高度就是最高的

        for (i = 0; i < 3*n; i++)
        {
            for (j = 0; j < i; j++)
            {
                if ((no[i].a < no[j].a && no[i].b < no[j].b) || (no[i].a < no[j].b && no[i].b < no[j].a)) ///因为储存的长和宽是不确定的
                    A[i] = max(A[i], A[j]+no[i].c);
            }
        } 

        for (i = 0; i < 3*n; i++)
            ans = max(ans, A[i]);

        printf("Case %d: maximum height = %d
", num, ans);
    }

    return 0;
}