专题三--1005

题目

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

思路

给出多种木块，每种木块有无数多个，求木块可垒成的最大高度。
每种木块有三种摆放方式，当一个木块有两个边都大于另一个木块时，两木块相容。
将每个木块分解为三个木块之后（三种摆放方式），就形成了一个类似于DAG上的最长路的问题（两个木块之间是二元关系且不会形成环）。使用动态规划来来考虑的话，状态转移方程是

dp[i]=max{dp[j]+b[i].height,j为与i块木块相容的木块的序号}

需要注意的几点是:

1. dp之前要先对木块进行排序，确保在进行外层的遍历时，与i相容的木块都在i之前，i之后不会出现于i相容的木块
2. 注意木块相容的判断条件，只要两条边都大即可，不必严格地宽大于宽，长大于长。

AC代码

1. #include<iostream>
2. #include<algorithm>
3. #include<stdio.h>
4. using namespace std;
5. struct block{
6. int c;
7. int k;
8. int g;
9. };
10. bool cmp(block a,block b){
11. if((a.c*a.k)<(b.c*b.k)){
12. return 1;
13. }
14. else return 0;
15. }
16. int MAX(int a,int b){
17. return a>b?a:b;
18. }
19. int main(){
20. int n;
21. struct block tem;
22. struct block a[300];
23. int dp[300];
24. int flag=0;
25. //freopen("date.in","r",stdin);
26. //freopen("date.out","w",stdout);
27. while(cin>>n&&n!=0){
28. flag++;
29. for(int i=1;i<=n;i++){
30. cin>>tem.c>>tem.k>>tem.g;
31. a[i*3].c=tem.c;a[i*3].k=tem.k;a[i*3].g=tem.g;
32. //a[i*6-1].c=tem.c;a[i*6-1].k=tem.g;a[i*6-1].g=tem.k;
33. a[i*3-1].c=tem.k;a[i*3-1].k=tem.g;a[i*3-1].g=tem.c;
34. //a[i*6-3].c=tem.k;a[i*6-3].k=tem.c;a[i*6-3].g=tem.g;
35. a[i*3-2].c=tem.g;a[i*3-2].k=tem.c;a[i*3-2].g=tem.k;
36. //a[i*6-5].c=tem.g;a[i*6-5].k=tem.k;a[i*6-5].g=tem.c;
37. }
38. sort(a+1,a+3*n+1,cmp);
39. /*for(int i=1;i<=3*n;i++){
40. cout<<a[i].c<<"\t"<<a[i].k<<"\t"<<a[i].g<<"\t"<<endl;
41. }*/
42. /*for(int i=1;i<=n;i++)
43. dp[i]=a[i].g;*/
44. for(int i=1;i<=3*n;i++){
45. dp[i]=a[i].g;
46. for(int j=i-1;j>=1;j--){
47. if(a[i].c>a[j].c&&a[i].k>a[j].k||a[i].c>a[j].k&&a[i].k>a[j].c)
48. dp[i]=MAX(dp[j]+a[i].g,dp[i]);
49. }
50. }
51. sort(dp+1,dp+3*n+1);
52. cout<<"Case "<<flag<<": maximum height = "<<dp[3*n]<<endl;
53. }
54. }

来自为知笔记(Wiz)