acm_猴子与箱子

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本文解析了一道经典的算法题目——构建最高塔。题目要求利用不同尺寸的长方体块搭建最高的塔，并确保上方的每个块在尺寸上严格小于下方的块。文章提供了详细的思路分析及实现代码。

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题目：

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

Sample Output

Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342

题意：输入一个数，然后输出几组数据，以三个为一组，是长方体的长宽高，无顺序，每一组数据的长方体都有无数个，一组数据假设2,3,6。。那么可以是2,3为长宽，亦可以2，6.。也可以3，6.。然后把长方体堆起来放，上面的长方体必须长和宽都小于下面的，求最大高度。。

想法：

可以先根据长或宽排序，然后求最长子序列。。

代码：

#include <iostream>
#include <algorithm>
using namespace std;
struct box1
{
int a,b,c;
int d;
};

bool cmp(box1 ss1,box1 ss2)
{
if(ss1.a<ss2.a) return 1;
else if(ss1.a==ss2.a&&ss1.b<ss2.b) return 1;
else return 0;
}
int inline max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int i,n,v=0;
box1 bo1[10001],bo2[10001];
//freopen("r.txt","r",stdin);
while(cin>>n&&n!=0)
{
v=v+1;
for(i=1;i<=n;i++)
{
cin>>bo1[i].a>>bo1[i].b>>bo1[i].c;
bo2[i].a=bo1[i].a;
bo2[i].b=bo1[i].b;
bo2[i].c=bo1[i].c;
bo2[i].d=bo1[i].c;
bo2[i+n].a=bo1[i].a;
bo2[i+n].b=bo1[i].c;
bo2[i+n].c=bo1[i].b;
bo2[i+n].d=bo1[i].b;
bo2[i+2*n].a=bo1[i].b;
bo2[i+2*n].b=bo1[i].c;
bo2[i+2*n].c=bo1[i].a;
bo2[i+2*n].d=bo1[i].a;
}
for(i=1;i<=3*n;i++)
{
int a;
if(bo2[i].a>bo2[i].b)
{
a=bo2[i].a;
bo2[i].a=bo2[i].b;
bo2[i].b=a;
}
}
sort(bo2+1,bo2+3*n+1,cmp);
int maxh=0;
for(i=2;i<=3*n;i++)
{
for(int j=0;j<i;j++)
if(bo2[i].a>bo2[j].a&&bo2[i].b>bo2[j].b)
bo2[i].d=max(bo2[j].d+bo2[i].c,bo2[i].d);
if(bo2[i].d>maxh) maxh=bo2[i].d;
}
cout<<"Case "<<v<<": maximum height = "<<maxh<<endl;
}
return 0;
}

做了好长时间。。