Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5829 Accepted Submission(s): 2961
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.
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.
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
题目的意思是说：给你了一个长方体的三个边，然后把大的放在下面，然后上面的依次小于下面的长和宽，并且每一个长方体都有无数个，
例如第一组例子：
可以这样放 最上面 第一层：20 10 30 分别是长，宽，高
下面放 第二层：30 20 10 这样最高的摆放就是40
其实这个题就是要求 最大子序列的题，不过要对长，宽，高先进行排序；每一组都有三种情况
1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 struct T 6 { 7 int x,y,z; 8 }c[100]; 9 int cmp(T a,T b) 10 { 11 if(a.x<b.x) 12 return 1; 13 if(a.x==b.x && a.y<b.y) 14 return 1; 15 return 0; 16 } 17 int main() 18 { 19 int i,j,n,t,max,a[4],k=0,opt[110]; 20 while(cin>>t && t) 21 {j=0;n=3*t;max=0; 22 memset(opt,0,sizeof(opt)); 23 while(t--) 24 { 25 for(i=0;i<3;i++) 26 cin>>a[i]; 27 sort(a,a+3);//如果不进行排序，下面就会有六种可能了， 28 c[j].x=a[2];c[j].y=a[1];c[j].z=a[0];j++;//a[2]>a[1]>a[0] 29 c[j].x=a[2];c[j].y=a[0];c[j].z=a[1];j++;//所以长始终大于或等于宽 30 c[j].x=a[1];c[j].y=a[0];c[j].z=a[2];j++;//这样就减少了三种情况 31 } 32 sort(c,c+n,cmp);//先按长进行排序，然后对宽进行排序 33 for(i=0;i<n;i++) 34 {opt[i]=c[i].z; 35 for(j=0;j<i;j++) 36 { 37 if(c[i].x>c[j].x && c[i].y>c[j].y && opt[j]+c[i].z>opt[i])//状态转移,选了就加上； 38 opt[i]=opt[j]+c[i].z; 39 } 40 // cout<<"opt="<<opt[i]<<endl; 41 } 42 for(i=0;i<n;i++) 43 if(opt[i]>max) 44 max=opt[i]; 45 k++; 46 cout<<"Case "<<k<<": maximum height = "<<max<<endl; 47 } 48 return 0; 49 }