C-Monkey and Banana HDU 1069 (Dynamic Planning + stacked cubes)

C-Monkey and Banana HDU 1069 (Dynamic Planning + stacked cubes)

C-Monkey and Banana

Time Limit: 1000 MS Memory Limit: 32768KB 64bit IO Format: % I64d & % I64u

Submit Status Practice HDU 1069

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 cocould 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 cocould 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 in 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

 

It was difficult to learn dynamic planning at the beginning. I have referred to the idea of this great god.

Dp [I]: Put the cube p [I] on the top to obtain the maximum height.

In this case, the status transfer is DP [I] = max (dp [I], p [I]. high + dp [j]).

 

#include
 
  #include#include
  
   #include
   
    #include
    
     #include
     
      using namespace std;template
      
       inline T read(T&x){ char c; while((c=getchar())<=32)if(c==EOF)return 0; bool ok=false; if(c=='-')ok=true,c=getchar(); for(x=0; c>32; c=getchar()) x=x*10+c-'0'; if(ok)x=-x; return 1;}template
       
         inline T read_(T&x,T&y){ return read(x)&&read(y);}template
        
          inline T read__(T&x,T&y,T&z){ return read(x)&&read(y)&&read(z);}template
         
           inline void write(T x){ if(x<0)putchar('-'),x=-x; if(x<10)putchar(x+'0'); else write(x/10),putchar(x%10+'0');}template
          
           inline void writeln(T x){ write(x); putchar('\n');}//-------ZCC IO template------const int maxn=400;const double inf=999999999;#define lson (rt<<1),L,M#define rson (rt<<1|1),M+1,R#define M ((L+R)>>1)#define For(i,t,n) for(int i=(t);i<(n);i++)typedef long long LL;typedef double DB;typedef pair
           
             P;#define bug printf("---\n");#define mod 1000000007struct node{ int width,length,area,high; bool operator<(const node&a) const { return area>a.area; }}p[maxn];int dp[maxn];int main(){ int n,m,i,j,t,k; int cas=1; while(read(n)&&n) { k=0; For(i,0,n) { int w,l,h; read__(w,l,h); node tmp; tmp.width=w;tmp.length=l;tmp.high=h; tmp.area=w*l; p[k++]=tmp; tmp.width=w;tmp.length=h;tmp.high=l; tmp.area=w*h; p[k++]=tmp; tmp.width=h;tmp.length=l;tmp.high=w; tmp.area=h*l; p[k++]=tmp; } sort(p,p+k); dp[0]=p[0].high; for(i=1;i
            
             p[i].width&&p[j].length>p[i].length||p[j].width>p[i].length&&p[j].length>p[i].width) { dp[i]=max(dp[i],dp[j]+p[i].high); } }// For(i,0,k)// printf("dp[%d] = %d\n",i,dp[i]); printf("Case %d: maximum height = %d\n",cas++,*max_element(dp,dp+k)); } return 0;}