HDU 1069 Monkey and Banana(DP)

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

写了很久的一道题。。调了很久的bug，最后发现问题在dp数组的初始化上。。。一开始应该把所有的h都读入dp，但是在之后的sort中，由于dp数组与cod无关，所以所有顺序都被打乱了，一些博客上将dp也写入结构体果然不无道理

注意立方体的排序，dp状态方程：选出长和宽都较小的，dp[i] = max(dp[j]+cod[i].h,dp[i])

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[200];
struct node{
	int l, w, h;
}cod[200];
bool cmp(node a, node b)
{
	if(a.l == b.l) return a.w < b.w;//更宽更长的放在下面 
	return a.l < b.l;
}
int main()
{
	std::ios::sync_with_stdio(false);
//	freopen("in.txt", "r", stdin);
	int n, t = 1;
	while(cin >> n)
	{
		if(!n)
			break;
		memset(dp, 0, sizeof(dp));
		int num = 0;
		int a, b, c;
		for(int i = 0; i < n; i++)
		{
			cin >> a >> b >> c;
			cod[num].l = a;cod[num].w = b;cod[num].h = c;dp[num] = cod[num++].h;
			cod[num].l = a;cod[num].w = c;cod[num].h = b;dp[num] = cod[num++].h;
			cod[num].l = b;cod[num].w = a;cod[num].h = c;dp[num] = cod[num++].h;
			cod[num].l = c;cod[num].w = a;cod[num].h = b;dp[num] = cod[num++].h;
			cod[num].l = c;cod[num].w = b;cod[num].h = a;dp[num] = cod[num++].h;
			cod[num].l = b;cod[num].w = c;cod[num].h = a;dp[num] = cod[num++].h;
		}
		sort(cod, cod+num, cmp);	//!!排序后dp顺序被打乱！！所以很多博客将dp也写入结构体 
		for(int i = 0; i < num; i++)
		{
			dp[i] = cod[i].h;
		}
//		dp[0] = cod[0].h;
		int max_h = 0;
		for(int i = 1; i < num; i++)
		{
//			max_h = 0;       /*更新max_h的情况：上一轮以j为最底层方块*/ 
			for(int j = 0; j < i; j++)
			{
				if(cod[j].l < cod[i].l&&cod[j].w < cod[i].w)//符合叠加条件 
				{
					dp[i] = max(dp[i], dp[j] + cod[i].h); 
//					max_h = max(dp[j], max_h);	/*与之前的max比较，择出最优max_h*/ 
				}
			}
			max_h = max(dp[i], max_h);
//			dp[i] = cod[i].h + max_h;	/*更新以i在最下面的dp*/ 
		}
		 cout<<"Case "<< t++ <<": maximum height = "<< max_h <<endl;
	}	
}