HDU

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

思路：

动规01背包题，得到每一个block当底的最大高度，最终输出最大值。

代码：

#include <cstdio>
#include <iostream>
#include <queue>
#include <algorithm>

using namespace std;

struct D{
	int x,y,z;
	int dp;
}b[200];

bool cmp(D a,D b){
	if(a.x!=b.x)return a.x<b.x;
	else if(a.y!=b.y)return a.y<b.y;
	else return a.z<b.z;
}

int main(){
	int N;
	int flag = 0;
	while(scanf("%d",&N)&&N){
		flag++;
		int k = 0;
		int x,y,z;
		while(N--){
			scanf("%d %d %d",&x,&y,&z);
			if(x==y)
            {
                if(y==z)
                {
                    b[k].x=x;b[k].y=y;b[k].z=z;b[k].dp=b[k].z;k++;
                }
                else  
                {
                    b[k].x=x;b[k].y=y;b[k].z=z;b[k].dp=b[k].z;k++;
                    b[k].x=z;b[k].y=y;b[k].z=x;b[k].dp=b[k].z;k++;
                    b[k].x=y;b[k].y=z;b[k].z=x;b[k].dp=b[k].z;k++;
                }                         
            }
            else
            {
                if(y==z)
                {
                    b[k].x=x;b[k].y=y;b[k].z=z;b[k].dp=b[k].z;k++;
                    b[k].x=y;b[k].y=x;b[k].z=z;b[k].dp=b[k].z;k++;
                    b[k].x=y;b[k].y=z;b[k].z=x;b[k].dp=b[k].z;k++;
                } 
                else if(x==z)
                {
                    b[k].x=x;b[k].y=y;b[k].z=z;b[k].dp=b[k].z;k++;
                    b[k].x=y;b[k].y=x;b[k].z=z;b[k].dp=b[k].z;k++;
                    b[k].x=x;b[k].y=z;b[k].z=y;b[k].dp=b[k].z;k++;
                }  
                else
                {
                    b[k].x=x;b[k].y=y;b[k].z=z;b[k].dp=b[k].z;k++;
                    b[k].x=y;b[k].y=x;b[k].z=z;b[k].dp=b[k].z;k++;
                    b[k].x=x;b[k].y=z;b[k].z=y;b[k].dp=b[k].z;k++;
                    b[k].x=z;b[k].y=x;b[k].z=y;b[k].dp=b[k].z;k++;
                    b[k].x=y;b[k].y=z;b[k].z=x;b[k].dp=b[k].z;k++;
                    b[k].x=z;b[k].y=y;b[k].z=x;b[k].dp=b[k].z;k++;
                }          
            }        
		}
		sort(b,b+k,cmp);
		int result = 0;
        for(int i=1 ; i<k ; i++)
        {
            for(int j=0 ; j<i ; j++)
               if(b[i].x>b[j].x && b[i].y>b[j].y)
                  b[i].dp=max(b[j].dp+b[i].z,b[i].dp);
            if(b[i].dp>result)result = b[i].dp;
        } 
        printf("Case %d: maximum height = %d
",flag,result);  
	}
	return 0;
}