1 題目編號：105
2 題目內容：
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.<br><br>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. <br><br>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. <br><br>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.<br>
Input
The input file will contain one or more test cases. The first line of each test case contains an integer n,<br>representing the number of different blocks in the following data set. The maximum value for n is 30.<br>Each of the next n lines contains three integers
representing the values xi, yi and zi.<br>Input is terminated by a value of zero (0) for n.<br>
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".<br>
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
3 題意：把給定的長方體（不限）疊加在一起，疊加的條件是，上面一個長方體的長和寬都比下面長方體的長和寬短；求這些長方體能疊加的最高的高度.（其中（3，2，1）可以擺放成（3，1，2）、（2，1，3）等）。
4 解題思路形成過程：每塊積木最多有3 個不同的底面和高度，我們可以把每塊積木看成三個不同的積木，那麼n種類型的積木就轉化爲3*n個不同的積木，對這3* n個積木的長按照從大到小排序，然後找到一個遞減的子序列，使得子序列的高度和最大。
5 代碼：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
struct node
{
int l, w, h;
} a[1047];
bool cmp(node a, node b)
{
if (a.l == b.l)
{
return a.w > b.w;
}
return a.l > b.l;
}
int MAX(int a, int b)
{
if (a > b)
return a;
return b;
}
int dp[1047];
int main()
{
int n;
int cas = 0;
while (scanf("%d", &n) && n)
{
//int L, W, H;
int tt[3];
int k = 0;
for (int i = 0; i < n; i++)
{
scanf("%d%d%d", &tt[0], &tt[1], &tt[2]);
sort(tt, tt + 3);
a[k].l = tt[0];
a[k].w = tt[1];
a[k].h = tt[2];
k++;
a[k].l = tt[1];
a[k].w = tt[2];
a[k].h = tt[0];
k++;
a[k].l = tt[0];
a[k].w = tt[2];
a[k].h = tt[1];
k++;
}
sort(a, a + k, cmp);
int maxx = 0;
for (int i = 0; i < k; i++)
{
dp[i] = a[i].h;
for (int j = i - 1; j >= 0; j--)
{
if (a[j].l>a[i].l && a[j].w>a[i].w)
{
dp[i] = MAX(dp[i], dp[j] + a[i].h);
}
}
if (dp[i] > maxx)
{
maxx = dp[i];
}
}
printf("Case %d: maximum height = %d\n", ++cas, maxx);
}
return 0;
}
