HDU 1069 Monkey and Banana（动态规划DP 经典）

原题

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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

题意

把给定的长方体（不限）叠加在一起，叠加的条件是，上面一个长方体的长和宽都比下面长方体的长和宽短；
求这些长方体能叠加的最高的高度.（其中（3，2，1）可以摆放成（3，1，2）、（2，1，3）等）.

涉及知识及算法

动态规划 类似于求最长上升子序列
初始化 dp[i]=box[i].h
状态转移方程 dp[i] = max(dp[i]，dp[j] + box[i].h)

代码

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

struct node
{
int l,w,h;
};
//存放长方体不同的摆放方式
node box[91];
//前大后小
bool cmp(node a,node b)
{
if(a.l>b.l)
{
return true;
}
else if(a.l==b.l&&a.w>b.w)
{
return true;
}
else
{
return false;
}
}

int main()
{
int c=1;
//freopen("in.txt","r",stdin);
int n;
while(scanf("%d",&n)&&n)
{
int d[3];
for(int i=0;i<n;i++)
{
scanf("%d%d%d",&d[0],&d[1],&d[2]);
sort(d,d+3);
//将长方体按不同的高度分为三类
box[3*i].h=d[0];box[3*i].l=d[2];box[3*i].w=d[1];
box[3*i+1].h=d[1];box[3*i+1].l=d[2];box[3*i+1].w=d[0];
box[3*i+2].h=d[2];box[3*i+2].l=d[1];box[3*i+2].w=d[0];
}
//排一下序使得前面的面积大致大于后面的
sort(box,box+n*3,cmp);
int dp[91];
for(int i=0;i<n*3;i++)
{
dp[i]=box[i].h;
}
//从面积第二小的开始摆
for(int i=n*3-2;i>=0;i--)
{
for(int j=i+1;j<n*3;j++)
{
if(box[i].l>box[j].l&&box[i].w>box[j].w&&dp[i]<dp[j]+box[i].h)
{
dp[i]=dp[j]+box[i].h;
}
}
}
int sumh=-999;
for(int i=0;i<n*3;i++)
{
if(sumh<dp[i])
{
sumh=dp[i];
}
}
printf("Case %d: maximum height = %d\n",c++,sumh);
}
return 0;
}

代码引自CSDN博主秦石秦草，链接http://blog.csdn.net/qinmusiyan/article/details/7986263，表示感谢。