题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1069
Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18739 Accepted Submission(s): 9967
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.
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.
现在研究人员怕香蕉挂太高，搞得猴子吃不到。问你最高能搭成多少高。这样他们好确定香蕉的位置。
dp[x]表示用上第x块木块时能搭的最高高度。
dp[i] = max(dp[i],dp[j]+a[i].z); j ∈[0,i-1];
边界条件（初始化）:dp[i] = a[i].z （因为每个方块最优解至少比他自身的高度要高）
记得在状态转移前要加上条件（if(a[j].x > a[i].x && a[j].y > a[i].y)）
#include<bits/stdc++.h>
using namespace std;
#define max_v 185
struct node
{
int x,y,z;
};
bool cmp(node a,node b)
{
return a.x*a.y>b.x*b.y;
}
struct node a[max_v];
int dp[max_v];
int main()
{
int c=,n;
while(~ scanf("%d",&n))
{
if(n==)
break;
for(int i=; i<*n;)
{
int x,y,z;
scanf("%d %d %d",&x,&y,&z);
a[i].x=x,a[i].y=y,a[i].z=z;
i++;
a[i].x=x,a[i].y=z,a[i].z=y;
i++;
a[i].x=y,a[i].y=x,a[i].z=z;
i++;
a[i].x=y,a[i].y=z,a[i].z=x;
i++;
a[i].x=z,a[i].y=y,a[i].z=x;
i++;
a[i].x=z,a[i].y=x,a[i].z=y;
i++;
}
sort(a,a+*n,cmp);
dp[]=a[].z;
//dp[i] 表示用上第i给木块能达到的最大高度
for(int i=; i<*n; i++)
{
dp[i]=a[i].z;
for(int j=; j<i; j++)
{
if(a[i].x<a[j].x&&a[i].y<a[j].y)
{
dp[i]=max(dp[i],dp[j]+a[i].z);
}
}
}
int t=;
for(int i=; i<*n; i++)
{
if(t<dp[i])
{
t=dp[i];
}
}
printf("Case %d: maximum height = %d\n",++c,t);
}
}