HDU1069 Monkey and Banana （动态规划）

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13503    Accepted Submission(s): 7117

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

【题目要求：】有n个长方体，分别有长，宽，高，分别用x，y，z表示。把这些长方体堆起来，求能堆的最大高度。并且一个长方体放在另一个长方体的条件是上面的那个长方体的长和宽小于下面的。而且每一个长方体有三种变换形式。

【思路：】根据题意可知，先求出所有可能变换的形式，然后先按照x进行排序，再按照y进行排序，对排序后的求最长子序列即可。

#include 
#include 
#include 
#include 
#include 
using namespace std;

struct node{
	int x,y,z,dp;
}P[205];

int cmp(node A, node B){
	if(A.x < B.x)
		return 1;
	else if(A.x == B.x && A.y < B.y)
			return 1;
	return 0;
}

int main(){
	int n,x,y,z,k,ans = 1;
	while(~scanf("%d",&n) && n){
		k = 0;
		for(int i = 0; i < n; i ++){
			scanf("%d%d%d",&x,&y,&z);
			if(x == y){
				if(y == z){
					P[k].x = x;	P[k].y = y;	P[k].z = z;	P[k].dp = P[k].z;k ++;
				}
				else{
					P[k].x = x;	P[k].y = y;	P[k].z = z;	P[k].dp = P[k].z;k ++;
					P[k].x = z;	P[k].y = y;	P[k].z = x;	P[k].dp = P[k].z;k ++;
					P[k].x = y;	P[k].y = z;	P[k].z = x;	P[k].dp = P[k].z;k ++;
				}
			}
			else{
				if(y == z){
					P[k].x = x;	P[k].y = y;	P[k].z = z;	P[k].dp = P[k].z;k ++;
					P[k].x = y;	P[k].y = x;	P[k].z = x;	P[k].dp = P[k].z;k ++;
					P[k].x = y;	P[k].y = z;	P[k].z = x;	P[k].dp = P[k].z;k ++;
				}
				else if(x == z){
					P[k].x = x;	P[k].y = y;	P[k].z = z;	P[k].dp = P[k].z;k ++;
					P[k].x = x;	P[k].y = z;	P[k].z = y;	P[k].dp = P[k].z;k ++;
					P[k].x = y;	P[k].y = x;	P[k].z = z;	P[k].dp = P[k].z;k ++;
				}
				else{
					P[k].x = x;	P[k].y = y;	P[k].z = z;	P[k].dp = P[k].z;k ++;
					P[k].x = x; P[k].y = z; P[k].z = y; P[k].dp = P[k].z;k ++;
					P[k].x = y;	P[k].y = z;	P[k].z = x; P[k].dp = P[k].z;k ++;
					P[k].x = y; P[k].y = x; P[k].z = z; P[k].dp = P[k].z;k ++;
					P[k].x = z; P[k].y = x; P[k].z = y; P[k].dp = P[k].z;k ++;
					P[k].x = z; P[k].y = y; P[k].z = x;	P[k].dp = P[k].z;k ++;
				}
			}		
		}
		sort(P,P + k,cmp);
		int Max = 0;
 		 for(int i = 1; i < k; i ++){
			for(int j = 0; j < i; j ++){
				if(P[i].x > P[j].x && P[i].y > P[j].y)
					P[i].dp = max(P[i].dp,P[j].dp + P[i].z); 
			}
			 if(P[i].dp>Max)Max=P[i].dp;
		}
		printf("Case %d: maximum height = %d\n",ans ++,Max);
	}	
	return 0;
}