猴子放箱子吃香蕉题（Problem ID：1069）

题址：http://acm.hdu.edu.cn/showproblem.php?pid=1069

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

思路：由题意可知，可以先将三个数作为不同的长宽高，将箱子分为不同的状态（长宽高不同），然后再按照cmp函数规则对不同箱子进行排序，按照题意，长越大的越可以作为最底部的箱子，通过调用STL库的sort函数进行排序，用dp[]数组记录箱子的高。

    注意，动态规划第一层是i从k-2开始，即从箱子长*宽第二小的开始扫，第二层j从i+1开始扫，找有没有更小的面能完全放在第一层确定的箱子上面构成最优解。每做一次能求出一个dp[i]。在这有一个重要的点，千万弄懂了，动规dp[i]数组在这里表示当前，取到i编号箱子，的最优解。

AC代码：

#include<iostream>
#include<stdio.h>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

int dp[111];

struct node
{
int l,w,h;
}box[111];

bool cmp(node a,node b)
{
if(a.l>b.l)	return true;
if(a.l==b.l && a.w>b.w)	return true;
return false;
}

int main()
{
int d[3],n,i,j,c=1,k,sumh;
while(scanf("%d",&n)!=EOF && n)
{
k=0;
//printf("此时的k为:%d\n",k);
for(i=0;i<n;i++)
{
scanf("%d%d%d",&d[0],&d[1],&d[2]);
sort(d,d+3);
box[k].l=d[2];box[k].w=d[1];box[k].h=d[0];k++;
box[k].l=d[2];box[k].w=d[0];box[k].h=d[1];k++;
box[k].l=d[1];box[k].w=d[0];box[k].h=d[2];k++;
}
//printf("此时的k为:%d\n",k);  //k代表有k种形态的箱子
sort(box,box+k,cmp);
for(i=0;i<k;i++)
{
dp[i]=box[i].h;
//printf("%d\t",dp[i]);
}
//putchar(10);
for(i=k-2;i>=0;i--)  //第一层先取一个面，下面第二层是表示，找更小的面能不能放上第一层那个面上面
{
for(j=i+1;j<k;j++)  //至于放不放上去是由if判断的，但是具体的放没放程序员是无法直接看出来的，但是结果第i个的最优解dp[i]是能知道的
{
if(box[i].l>box[j].l && box[i].w>box[j].w)
if(dp[i]<dp[j]+box[i].h)
dp[i]=dp[j]+box[i].h;
}
}
sumh=dp[0];
for(i=0;i<k;i++)
if(sumh<dp[i])  //注意：dp[i]表示当前取到i编号这个箱子的最优解，即能叠起来的最高的高度
sumh=dp[i];
printf("Case %d: maximum height = %d\n",c++,sumh);
}

return 0;
}