题目连接：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): 10875 Accepted Submission(s): 5660
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.
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
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342
题目：给出一些长方体，然后让你把他堆成塔，
要求下面的塔的要比上面的塔大（长和宽），
而且每一种长方体的数量都是无限的。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = ; struct Node {
int x;
int y;
int z;
bool operator < (const Node &a) const
{
if(x!=a.x) return x < a.x;
else if(y!=a.y) return y < a.y;
else return z > a.z;
}
} node[N];
int dp[N]; int main()
{
int n;
int cnt = ;
while(~scanf("%d",&n))
{
if(n==) return ;
memset(dp,,sizeof(dp));
int x,y,z;
int t = ;
for(int i = ; i < n; i++){
scanf("%d%d%d",&x,&y,&z);
node[t].x = x;
node[t].y = y;
node[t].z = z;
t++;
node[t].x = x;
node[t].y = z;
node[t].z = y;
t++;
node[t].x = y;
node[t].y = x;
node[t].z = z;
t++;
node[t].x = y;
node[t].y = z;
node[t].z = x;
t++;
node[t].x = z;
node[t].y = x;
node[t].z = y;
t++;
node[t].x = z;
node[t].y = y;
node[t].z = x;
t++;
}
sort(node,node+*n);
int mmax = ;
for(int i = ; i < *n; i++)
dp[i] = node[i].z;
for(int i = ; i < *n; i++)
{
for(int j = ; j < i; j++)
{
if((node[i].x>node[j].x)&&(node[i].y>node[j].y))
dp[i] = max(dp[i],dp[j]+node[i].z);
}
mmax = max(mmax,dp[i]);
}
printf("Case %d: maximum height = ",cnt);
cnt++;
printf("%d\n",mmax);
}
return ;
}