| Time limit | 1000 ms | Memory limit | 32768 kB |
题目：
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.
输入：
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.
输出：
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".
样例输入：
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
样例输出：
Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342
题意：给出一个数字n，然后输入n个立体块的长宽高，每个立体块是无限的，现在想把这些立体块搭建起来，一个立体块能放在另外一个立体块上的原则是上面的立体块的长和宽都要比下面的立体块小，问利用这些立体块能搭建的最大高度。
思路：把每一个输入的立体块的长宽高三个参数分别调换顺序，保证这个立体块的每个面都能作为底面进行搭建，这就是列举出了所有能够搭建的立体块，然后对所有的立体块的长进行排序，然后建立dp数组，dp[i]表示以第i个立体块为最下面的立体块能搭建的最大高度，那么dp[i]=max{能放在第i个立体块上的所有立体块的dp}+第i个立体块自身的高度。最后在找一遍最大值。
AC代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[2000];
struct node
{
int x;int y;int z;
}a[2000];
bool cmp(node m,node n)
{
return m.x<n.x;
}
int main()
{
int n,cont,Max,x1,y1,z1,g=0;
while(cin>>n)
{
if(n==0)
break;
memset(dp,0,sizeof(dp));
cont=-1;
for(int i=1;i<=n;i++)
{
scanf("%d %d %d",&x1,&y1,&z1);
a[++cont].x=x1;a[cont].y=y1;a[cont].z=z1;
a[++cont].x=x1;a[cont].y=z1;a[cont].z=y1;
a[++cont].x=y1;a[cont].y=x1;a[cont].z=z1;
a[++cont].x=y1;a[cont].y=z1;a[cont].z=x1;
a[++cont].x=z1;a[cont].y=x1;a[cont].z=y1;
a[++cont].x=z1;a[cont].y=y1;a[cont].z=x1;
}
sort(a,a+cont,cmp);
dp[0]=a[0].z;
for(int i=0;i<=cont;i++)
{
Max=0;
for(int j=i-1;j>=0;j--)
{
if((a[j].x<a[i].x)&&(a[j].y<a[i].y)&&(dp[j]>Max))
Max=dp[j];
}
dp[i]=a[i].z+Max;
}
printf("Case %d: maximum height = %d
",++g,*max_element(dp,dp+cont));
}
}注意点：最后要求出以所有立体块底面所能搭建的高度的最大值。