Monkey and Banana(dp)

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. 

InputThe 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. 
OutputFor 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

dp,因为一个输入长方形其实不止一种长宽高，所以先记录所有可能，然后排序，从顶到下搭塔，如果下一块比上一块大，就可以搭上去，dp记录最优高度

#include<stdio.h>
#include<algorithm>
using namespace std;
int n,a,b,c,dp[33];
struct s{
    int l,w,h;//定义长宽高结构体 
}x[200];

bool cmp(s a,s b){//按长从小到大，如果一样，宽从小到大排序 
    if(a.l==b.l)
        return a.w<b.w;
    return a.l<b.l;
}

int main(){
    int k=1;
    while(scanf("%d",&n)!=EOF&&n){
        int cnt=0;
        for(int i=0;i<n;i++){
            scanf("%d%d%d",&a,&b,&c);
            x[cnt].h=a,x[cnt].l=b,x[cnt++].w=c;//长方形的6种摆法 
            x[cnt].h=a,x[cnt].l=c,x[cnt++].w=b;
            x[cnt].h=b,x[cnt].l=a,x[cnt++].w=c;
            x[cnt].h=b,x[cnt].l=c,x[cnt++].w=a;
            x[cnt].h=c,x[cnt].l=a,x[cnt++].w=b;
            x[cnt].h=c,x[cnt].l=b,x[cnt++].w=a;
        }
        sort(x,x+cnt,cmp);//排序 
        for(int i=0;i<cnt;i++){
            dp[i]=x[i].h;
            for(int j=0;j<i;j++){//从j到i的最大高度 
                if(x[j].l<x[i].l&&x[j].w<x[i].w)//从顶到下，也就是上面的长方形要比下面的长方形的长宽小 
                    dp[i]=max(dp[i],dp[j]+x[i].h);//找到i之前能堆起来的最高高度，加上i的高度，都不符合，dp[i]就是i的高度 
            }
        }
        sort(dp,dp+cnt);//排序找到最高的dp 
        printf("Case %d: maximum height = %d
",k++,dp[cnt-1]);//一个pe的地方，等号前后都有空格 
    }
    return 0;
}