杭电 1069 Monkey and Banana

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

经典的dp问题，刚开始接触有点难度

给出n种类型的长方体石块，给出每种的长宽高，每种石块有三种摆放方式，数量不限，可以累加着往上放，问最多能放多高

但是有个限定条件，长和宽必须均小于下面石块的长和宽才能放上去

转换成dp模型，就是dp[i]表示放置到第i个石块最高的高度，上面石块边长必须严格小于下面石块

 #include<cstdio>
 #include<algorithm>
 using namespace std;
 struct stu
 {
     int l,w,h;
 }st[];
 bool cmp(stu a,stu b)                        //从顶往下判断所以从小往大排序
 {
     if(a.l != b.l)
         return a.l < b.l;
     else
         return a.w < b.w;
 }
 int main()
 {
     int k=;
     int dp[];                            //dp[i]表示从顶开始到第i个木块的高度
     int n,a,b,c,max0,num;
     while(scanf("%d",&n) && n)
     {
         int i,j;
         num=;
         for(i =  ; i <= n ; i++)
         {
             scanf("%d %d %d",&a,&b,&c);
             st[num].l=a,st[num].w=b,st[num++].h=c;        //每个木块有三种放法
             st[num].l=a,st[num].w=c,st[num++].h=b;
             st[num].l=b,st[num].w=a,st[num++].h=c;
             st[num].l=b,st[num].w=c,st[num++].h=a;
             st[num].l=c,st[num].w=a,st[num++].h=b;
             st[num].l=c,st[num].w=b,st[num++].h=a;
         }
         sort(st,st+num,cmp);
         for(i =  ; i < num ; i++)
         {
             dp[i]=st[i].h;
         }
         for(i =  ; i < num ; i++)
         {
             for(j =  ; j < i ; j++)
             {
                 if(st[j].l < st[i].l && st[j].w < st[i].w && dp[j]+st[i].h > dp[i])
                 {
                     dp[i]=dp[j]+st[i].h;
                 }
             }
         }
         sort(dp,dp+num);
         printf("Case %d: maximum height = ",k++);
         printf("%d\n",dp[num-]);
     }
     return ;
 }