專題三--1005

題目

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

給出多種木塊，每種木塊有無數多個，求木塊可壘成的最大高度。
每種木塊有三種擺放方式，當一個木塊有兩個邊都大于另一個木塊時，兩木塊相容。
將每個木塊分解為三個木塊之后（三種擺放方式），就形成了一個類似于DAG上的最長路的問題（兩個木塊之間是二元關系且不會形成環）。使用動態規劃來來考慮的話，狀態轉移方程是

dp[i]=max{dp[j]+b[i].height,j為與i塊木塊相容的木塊的序號}

需要注意的幾點是:

1. dp之前要先對木塊進行排序，確保在進行外層的遍歷時，與i相容的木塊都在i之前，i之后不會出現于i相容的木塊
2. 注意木塊相容的判斷條件，只要兩條邊都大即可，不必嚴格地寬大于寬，長大于長。

AC代碼

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. }

來自為知筆記(Wiz)