Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10987 Accepted Submission(s): 5732
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
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：叠长方体，要求上面的长方体地面长宽都要小于下面的长和宽，求能叠的最大高度。
思路：可知上面的面积要比下面的小，一个长方体有三种摆放位置，可将所有的长方体，所有摆放状态进行按照面积排序，这样必然摆放的序列必然是他的子序列，转化为求最长上升子序列问题。需要注意的是宽和长一定要宽<=长，这样能够保证能够一一对应上去。
状态转移方程：dp[i]=max(dp[i],dp[j]+a[i].z | a[j].x<a[i].x&&a[j].y<a[i].y)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 struct blo 7 { 8 int x,y,z; 9 }a[95]; 10 int dp[95]; 11 bool cmp(blo u,blo v) 12 { 13 int t=u.x*u.y-v.x*v.y; 14 if(t==0) 15 return u.z<=v.z; 16 else if(t<0) 17 return 1; 18 else 19 return 0; 20 } 21 int main() 22 { 23 int i,j; 24 int n; 25 int t=1; 26 freopen("in.txt","r",stdin); 27 while(scanf("%d",&n)&&n) 28 { 29 memset(dp,0,sizeof(dp)); 30 for(i=1;i<=3*n;i+=3) 31 { 32 scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z); 33 if(a[i].x>a[i].y) 34 { 35 int temp=a[i].x; 36 a[i].x=a[i].y; 37 a[i].y=temp; 38 } 39 a[i+1].x=min(a[i].y,a[i].z),a[i+1].y=max(a[i].y,a[i].z),a[i+1].z=a[i].x; 40 a[i+2].x=min(a[i].x,a[i].z),a[i+2].y=max(a[i].x,a[i].z),a[i+2].z=a[i].y; 41 } 42 sort(a+1,a+3*n+1,cmp); 43 for(i=1;i<=3*n;i++) 44 { 45 dp[i]=a[i].z; 46 for(j=1;j<i;j++) 47 { 48 if(a[j].x<a[i].x&&a[j].y<a[i].y&&(dp[j]+a[i].z)>dp[i]) 49 dp[i]=dp[j]+a[i].z; 50 } 51 } 52 int m=0; 53 for(i=1;i<=3*n;i++) 54 m=max(m,dp[i]); 55 printf("Case %d: maximum height = %d ",t++,m); 56 } 57 }