- Monkey and Banana



Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  

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".

         
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 

Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342 

开始不知道怎么解决无后效性问题，后来想了想排序，觉得不科学，后来问了下学长 ，真的是排序。
最大递减子序列问题，代码如下：
 

//
// Create by 神舟 on 2015-02-11
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <cctype>
#include <stack>
#include <queue>
#include <map>
#include <string>
#include <set>
#include <vector>
using namespace std;

#define CLR(x) memset(x,0,sizeof x)
#define ll long long
#define inf 0x3f3f3f3f
const int maxn=1e3+5;
const int MOD=1e5+5;
int k,cas=0,dp[maxn];
struct nnn
{
    int x,y,h;
}num[maxn];
bool cmp(nnn x,nnn y){
    return x.x*x.y>y.x*y.y;
}
int main()
{
#ifdef LOCAL
 freopen("in.txt","r",stdin);
 //freopen("out.txt","w",stdout);
#endif
 ios_base::sync_with_stdio(0);

 int n;
 while(++cas&&scanf("%d",&n)!=EOF&&n){
        int x,y,z;
        k=1;
        memset(dp,0xc3,sizeof dp);

        for(int i=1;i<=n;i++){
            scanf("%d%d%d",&x,&y,&z);
            num[k].x=y,num[k].y=z,num[k].h=x,k++;
            num[k].x=x,num[k].y=y,num[k].h=z,k++;
            num[k].x=z,num[k].y=x,num[k].h=y,k++;
        }

        sort(num+1,num+k,cmp);
        int ans=-inf;
        dp[0]=0;
        num[0].x=num[0].y=inf,num[0].h=0;
        k--;
        for(int i=1;i<=k;i++) for(int j=i-1;j>=0;j--)
            if((num[i].x<num[j].x&&num[i].y<num[j].y)||(num[i].y<num[j].x&&num[i].x<num[j].y)){
                dp[i]=max(dp[i],dp[j]+num[i].h);
        }
        for(int i=1;i<=k;i++) ans=max(ans,dp[i]);
        printf("Case %d: maximum height = %d\n",cas,ans);
 }

 return 0;
}