hdu1069之Monkey and Banana

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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    简述题意：输入n代表n个箱子，分别有接下来n行有x,y,z代表第i个箱子的长宽高，每个种箱子有无限个并且可以旋转方向放，一个箱子可以放到另一个箱子上面，条件是上面的箱子相应的长宽比下面箱子相应的长宽小，问箱子叠起来最高有多少。
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;

const int MAX=100;
int height[MAX];

typedef struct node{
	int w;
	int l;
	int h;
	bool operator<(node b)const{
		if(w == b.w)return l>b.l;
		return w>b.w;
	}
};
node s[MAX];

int main(){
	int n,a,b,c,num=1;
	while(cin>>n,n){
		int k=0;
		for(int i=0;i<n;++i){
			cin>>a>>b>>c;
			//对a,b,c排序. 
			if(a>b)a=a+b,b=a-b,a=a-b;
			if(a>c)a=a+c,c=a-c,a=a-c;
			if(b>c)b=b+c,c=b-c,b=b-c;
			s[k].w=a,s[k].l=b,s[k++].h=c;//以a,b为底,c为高. 
			s[k].w=a,s[k].l=c,s[k++].h=b;//以a,c为底,b为高.
			s[k].w=b,s[k].l=c,s[k++].h=a;//以b.c为底,a为高. 
		}
		sort(s,s+k);//排序,以底边中最小的边来排序,这样使得第i个箱子只能放在第i个箱子的前面的箱子上,当然也可以按照底边面积排序. 
		int Max=s[0].h;
		height[0]=s[0].h;
		for(int i=1;i<k;++i){
			height[i]=s[i].h;
			for(int j=0;j<i;++j){
				if(s[i].w<s[j].w && s[i].l<s[j].l){
					height[i]=max(height[i],s[i].h+height[j]);
				}
			}
			Max=max(Max,height[i]);
		}
		cout<<"Case "<<num++<<": maximum height = "<<Max<<endl;
	}
	return 0;
}