Problem E

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

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

题意：给你几种类型的积木，每种类型的无数个，可以随便用，让你求最高能摞多高，要求底下的一定比上面的面积大；

解题思路：面积无非就是长宽，在长排序一定的条件下，对宽求最长递减子序列，比赛前攒出来的题但是案例没通过，回来想了一会，刚开始想的是求最长递增子序列，不知哪里错了，后来又改成最长递减子序列，结果排序函数有没改，愁人啊；

感悟：一道题不能间隔太长时间啊，回来啥都忘了，还得看自己代码，仔细回忆。。。。。

代码：

#include
#include
#include
#include
#define maxn 200
using namespace std;
struct node
{
    int x,y,z;
    bool operator <(node const &other) const
    {
            return this->x
    }//将长按照从大到小的顺序排好
};
node ans[maxn];
int n,x[4];
int main()
{
    //freopen("in.txt","r",stdin);
    int c=0,d=0,dp[maxn];
    while(~scanf("%d",&n)&&n)
    {
        memset(dp,0,sizeof dp);
        d=1;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d",&x[0],&x[1],&x[2]);
            sort(x,x+3);
            ans[d].x=x[0],ans[d].y=x[1],ans[d++].z=x[2];
            ans[d].x=x[0],ans[d].y=x[2],ans[d++].z=x[1];
            ans[d].x=x[1],ans[d].y=x[2],ans[d++].z=x[0];
        }//输入结果
        sort(&ans[1],&ans[3*n+1]);
        for(int i=1;i<=3*n;i++)
            dp[i]=ans[i].z;
        dp[0]=0;
        ans[0].x=ans[0].y=ans[0].z=0;
        //for(int i=0;i<=6*n;i++)
            //cout<<"ans[i].y="<<ans[i].y<<" "<<"ans[i].z="<<ans[i].z<<endl;
        //cout<<endl;
        int amax=-1;
        for(int i=1;i<=3*n;i++)//对宽进行求最长递增子序列
        {
            int m=-1;
            for(int j=i-1;j>=0;j--)
            {
                if(ans[i].x>ans[j].x&&ans[i].y>ans[j].y)
                {
                    m=max(dp[j],m);
                }
            }
            dp[i]+=m;
            //cout<<dp[i]<<endl;
            if(amax
                amax=dp[i];
        }
        printf("Case %d: maximum height = %d ",++c,amax);
    }
}