Monkey and Banana 【HDU】-1069

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Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15952    Accepted Submission(s): 8463

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

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

 

Source

University of Ulm Local Contest 1996

 

中文版：點擊打開鏈接

計算猴子們最高可以堆出的磚塊們的最大高度。

代碼：

#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
#include <vector>
using namespace std;
#define INF 0x7fffffff 
struct Brick
{
	int x,y,z;
}brick[200];
int ant;
void UpDate(int x,int y,int z)   //每個磚塊的3條邊可以組成6種不同的長寬高
{
	brick[ant].x=x;
	brick[ant].y=y;
	brick[ant++].z=z;
	
	brick[ant].x=x;
	brick[ant].y=z;
	brick[ant++].z=y;
	
	brick[ant].x=y;
	brick[ant].y=x;
	brick[ant++].z=z;
	
	brick[ant].x=y;
	brick[ant].y=z;
	brick[ant++].z=x;
	
	brick[ant].x=z;
	brick[ant].y=x;
	brick[ant++].z=y;
	
	brick[ant].x=z;
	brick[ant].y=y;
	brick[ant++].z=x;
}
bool cmp(Brick a,Brick b)   //長或寬必須是逐漸減小 
{
	if(a.x==b.x)
	return a.y>b.y;
	return a.x>b.x; 
}
int main()
{
	int n,Case=1;
	while(scanf("%d",&n)!=EOF&&n)
	{
		ant=1;
		for(int i=1;i<=n;i++)
		{
			int x,y,z;
			scanf("%d%d%d",&x,&y,&z);
			UpDate(x,y,z);
		}
		ant--;
		sort(brick+1,brick+ant+1,cmp);
		
		int dp[200],ans=0;
		for(int i=1;i<=ant;i++)
		{
			dp[i]=brick[i].z;   //第一塊磚的高度 
			for(int j=1;j<i;j++)
			{
				if(brick[i].x < brick[j].x && brick[i].y < brick[j].y)  //滿足題意下面的磚頭比在它上面的磚頭小 
				dp[i]=max(dp[i],dp[j]+brick[i].z);  // dp[j]表示dp[j:(1~i-1)]以第i-1塊磚的高度和 
				ans=max(ans , dp[i]);    
				
			}
		}
		printf("Case %d: maximum height = %d\n",Case++,ans);
	}
	return 0;
}
//dp[i]:  i表示以第i塊磚頭結尾能到達的最高高度 
//dp[i] = max(brick[i].z , dp[j:(1~i-1)] + brick[i].z)
//if : brick[i].x < brick[j].x && brick[i].y < brick[j].y