一 题意描述：
Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3291 Accepted Submission(s): 1706
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 isclever 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. Theproblem is that, in building a tower, one block could only be placed on top of another block as long asthe two base dimensions of the upper block were both strictly smaller than the corresponding basedimensions of the lower block because there has to be some space for the monkey to step on. Thismeant, 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. Theproblem is that, in building a tower, one block could only be placed on top of another block as long asthe two base dimensions of the upper block were both strictly smaller than the corresponding basedimensions of the lower block because there has to be some space for the monkey to step on. Thismeant, 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
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
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342
二 思路分析：
对于本题，我们首先可以确定一个大致方向，就是首先要放底面积大的方块。而对于每一个方块，我们有三种方法，那么我们可以将每种方块的三个状态存放起来，并按照面积大小从大到小进行排序。完成开头工作后，下面我们进行最重要的工作---建立dp方程：我们设dp[i]表示以第i块砖作为整座塔的塔顶时整座塔的最大高度。
那么dp[i]=max{dp[j]+ret[i].z | 0<=j<i,(ret[j].x>ret[i].x&&ret[j].y>ret[i].y)||(ret[j].y>ret[i].x&&ret[j].x>ret[i].y) }.到这里我们就可以编写代码了。
三 AC代码：
1 #include<iostream> 2 #include <algorithm> 3 using namespace std; 4 //定义结构体 5 struct Rectangle 6 { 7 int x,y,z; 8 int s; 9 }; 10 Rectangle ret[100]; 11 int dp[100]; 12 int cmp(Rectangle a,Rectangle b)//将面积按照降序排列 13 { 14 return a.s>b.s; 15 } 16 int main() 17 { 18 int n; 19 int x,y,z; 20 int num=1; 21 while(cin>>n) 22 { 23 if(n==0)break; 24 int i=0; 25 for(int j=0;j<n;j++)//输入数据并转化为三种形式存放 26 { 27 cin>>x>>y>>z;//输入x,y，z 28 ret[i].x=x;//把z作为高 29 ret[i].y=y; 30 ret[i].z=z; 31 ret[i].s=ret[i].x*ret[i].y; 32 ret[i+1].x=z;//把x作为高 33 ret[i+1].y=y; 34 ret[i+1].z=x; 35 ret[i+1].s=ret[i+1].x*ret[i+1].y; 36 ret[i+2].x=x;//把y作为高 37 ret[i+2].y=z; 38 ret[i+2].z=y; 39 ret[i+2].s=ret[i+2].x*ret[i+2].y; 40 i+=3; 41 } 42 sort(ret,ret+3*n,cmp);//排序 43 for(int i=0;i<3*n;i++)//赋初值为本身的高度 44 dp[i]=ret[i].z; 45 for(int i=1;i<3*n;i++) 46 { 47 for(int j=0;j<i;j++) 48 if((ret[j].x>ret[i].x&&ret[j].y>ret[i].y)||(ret[j].y>ret[i].x&&ret[j].x>ret[i].y)) 49 if(dp[j]+ret[i].z>dp[i]) dp[i]=dp[j]+ret[i].z;//dp方程关键所在 50 } 51 int max_height=dp[0]; 52 for(int j=0;j<3*n;j++) 53 if(dp[j]>max_height) 54 max_height=dp[j]; 55 cout<<"Case"<<' '<<num<<':'<<" maximum height = "<<max_height<<endl; 56 num++; 57 } 58 return 0; 59 }