ZOJ Problem Set - 1093 Monkey and Banana

ZOJ Problem Set - 1093

Monkey and Banana

---

Time Limit: 2 Seconds                                    Memory Limit:65536 KB                            

---

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 (x_i, y_i, z_i).  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 Specification

The input file will contain one or more test cases. The first line of each test case contains an integern,
representing the number of different blocks in the following data set. The maximum value forn is 30.
Each of the next n lines contains three integers representing the valuesx_i,y_i andz_i.
Input is terminated by a value of zero (0) for n. 

Output Specification

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 "Casecase: 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

贪心算法，只是砖块有点麻烦

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
struct Box{
	int len,wid,hei;
};
Box box[600];
bool cmp(Box a,Box b)
{
	if(a.len!=b.len)
	{
		return a.len>b.len;
	}
	else return a.wid>b.wid;
}
void oriente(int index,int x,int y,int z)
{
	box[index].len=x;
	box[index].wid=y;
	box[index].hei=z;
}
int main()
{
	int n;
	int cas=0;
	while(cin>>n&&n!=0)
	{
		int number=0;
		for(int i=0;i<n;i++)
		{
			int t1,t2,t3;
			cin>>t1>>t2>>t3;
			oriente(number++,t1,t2,t3);
			oriente(number++,t1,t3,t2);
			oriente(number++,t2,t1,t3);
			oriente(number++,t2,t3,t1);
			oriente(number++,t3,t1,t2);
			oriente(number++,t3,t2,t1);
		}
		sort(box,box+number,cmp);
		int height[600];
		for(int i=0;i<number;i++)
		   height[i]=box[i].hei;
		int result=0;
		for(int i=1;i<number;i++)
		{
		    int max=0;
			for(int j=i-1;j>=0;j--)
			{
				if(box[j].len>box[i].len&&box[j].wid>box[i].wid&&max<height[j])
				{
				  max=height[j];
				}
			}
			height[i]+=max;
			if(result<height[i])
			   result=height[i];
		}
		cout<<"Case "<<++cas<<": maximum height = "<<result<<endl;
	}
}