【題意】：

有人要做實驗，看看猴子的智商，在屋頂上邊掛著吃的東西，然後給你一些木塊的長、寬、高，問你猴子最高可以爬多高，下面的木塊必須大於（不能等於）上面的木塊。

【思考】：

感覺像貪心，但還要多一點dp的味道，如果你可以看出這是一個嚴格遞增子序列的問題，那最好了，可惜我沒看出來，所以用了dp，dp轉移方程為：dp[i]=max( dp[i]，dp[j]+hi[i]) ，所以開始碼起來  ^_^ ！

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

ac程式碼：

記得長寬高要從小到大排一下序！！！

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

int n;
int dp[3000];
int a,b,c;

struct node{
	int l1,l2,hg;
}blk[3000];

bool cmp(node a,node b)
{
	if(a.l1>b.l1)	return true;
	if(a.l1==b.l1&&a.l2>b.l2 )	return true;
	else	return false;
}

int main()
{
	int ans,i,j,m=0,x,y,z;
	while(scanf("%d",&n)&&n)
	{
		memset(dp,0,sizeof(dp));
		int k=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			x=a,y=b,z=c;
			a=min(x,min(y,z));
			c=max(x,max(y,z));
			int p=x+y+z;
			b=p-a-c;
			blk[k].l1 =a, blk[k].l2 =b; blk[k].hg=c;
			k++;
			blk[k].l1 =a, blk[k].l2 =c; blk[k].hg=b;
			k++;
			blk[k].l1 =b, blk[k].l2 =a; blk[k].hg=c;
			k++;
			blk[k].l1 =b, blk[k].l2 =c; blk[k].hg=a;
			k++;
			blk[k].l1 =c, blk[k].l2 =b; blk[k].hg=a;
			k++;
			blk[k].l1 =c, blk[k].l2 =a; blk[k].hg=b;
		}
		sort(blk,blk+k,cmp);
		ans=0;
		for(i=k-1;i>=0;i--){
			dp[i]=blk[i].hg;
			for(j=i+1;j<k;j++){
				if(blk[i].l1 >blk[j].l1 && blk[i].l2 >blk[j].l2)
				{
					dp[i]=max(dp[i],dp[j]+blk[i].hg);
				}	
			}
		}
		for(i=0;i<k;i++)	
			ans=max(ans,dp[i]);
		printf("Case %d: maximum height = %d\n",++m,ans);
	}
	return 0;
}