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

 

Sample Output

 #include<bits/stdc++.h>

 using namespace std;

 int dp[];

 struct node

 {

     int x,y,z;

 }a[];

 bool cmp(node a,node b)//x升序  x相等  y升序

 {

     if(a.x!=b.x)

     {

         return a.x<b.x;

     }

     else

     {

         return a.y<=b.y;

     }

 }

 int main() {

     int n;int cases=;

     while(~scanf("%d",&n),n)

     {

         cases++;

         int num=;

         for(int i=;i<n;i++)

         {

             int x,y,z;

             scanf("%d %d %d",&x,&y,&z);

 //            if(x==y&&y==z)

 //            {

 //                a[num].x=x;

 //                a[num].y=x;

 //                a[num].z=x;

 //                num++;

 //            }

 //            else

 //            {

                 a[num].x=x;

                 a[num].y=y;

                 a[num].z=z;

                 num++;

                 a[num].x=x;

                 a[num].y=z;

                 a[num].z=y;

                 num++;

                 a[num].x=y;

                 a[num].y=x;

                 a[num].z=z;

                 num++;

                 a[num].x=y;

                 a[num].y=z;

                 a[num].z=x;

                 num++;

                 a[num].x=z;

                 a[num].y=x;

                 a[num].z=y;

                 num++;

                 a[num].x=z;

                 a[num].y=y;

                 a[num].z=x;

                 num++;

         //    }

         }

         memset(dp,,sizeof(dp));

         sort(a,a+num,cmp);

         int maxx=;

         dp[]=a[].z;

         for(int i=;i<num;i++)//每次都和前面的比    摆放要求上小下大，这边是从上往下找的，先最小的（最上面），再大的

         {

             int temp=;//记录前面几个里面高度最大的那个  因为每次都要最优，那么就要把他放在已有的高度最高那

             for(int j=i-;j>=;j--)

             {

                 if(a[i].x>a[j].x&&a[i].y>a[j].y&&dp[j]>temp)//因为排过序，所以大的在后，小的在前

                 {

                     temp=dp[j];

                 }

             }

             dp[i]=temp+a[i].z;//i位置处的高度

             maxx=max(dp[i],maxx);//所求的答案

         }

         printf("Case %d: maximum height = %d\n",cases,maxx);

     }

     return ;

 }