HDU 1069 Monkey and Banana (DP)

Monkey and Banana
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
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Practice  
HDU 1069
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. 
 
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
 
 
 
사고방식: 우선, 각 네모난 블록의 세 차원은 모두 3*2=6의 조합이 있는데 제목은 최대 30개의 네모난 블록이 있기 때문에 수조는 180이상 열어야 한다.사고방식은DP수조의 모든 원소가 이 네모난 블록을 밑으로 하는 최고 진동의 높이를 저장하고 맨 끝에 있는 것부터 표를 작성하는 동시에 최대치를 기록하고 마지막에 최대치를 출력하면 된다는 것이다.

 1 #include<stdio.h>

 2 #include<stdlib.h>

 3 #include<string.h>

 4 #define    MAX    200

 5 

 6 struct    x

 7 {

 8     int    x;

 9     int    y;

10     int    z;

11 }DP[MAX];

12 

13 int    comp(const void * a,const void * b);

14 int    main(void)

15 {

16     int    n,max,box,x,y,z,count;

17     count = 0;

18 

19     while(scanf("%d",&n) && n)

20     {

21         count ++;

22         for(int i = 0;i < 6 * n;i ++)

23         {

24             scanf("%d%d%d",&x,&y,&z);

25             DP[i].x = x;

26             DP[i].y = y;

27             DP[i].z = z;

28             

29             i ++;

30             DP[i].x = x;

31             DP[i].y = z;

32             DP[i].z = y;

33 

34             i ++;

35             DP[i].x = z;

36             DP[i].y = y;

37             DP[i].z = x;

38 

39             i ++;

40             DP[i].x = z;

41             DP[i].y = x;

42             DP[i].z = y;

43 

44             i ++;

45             DP[i].x = y;

46             DP[i].y = x;

47             DP[i].z = z;

48 

49             i ++;

50             DP[i].x = y;

51             DP[i].y = z;

52             DP[i].z = x;

53         }

54         qsort(DP,6 * n,sizeof(struct x),comp);

55 

56         max = DP[6 * n - 1].z;

57         for(int i = 6 * n - 2;i >= 0;i --)

58         {

59             box = 0;

60             for(int j = i + 1;j < 6 * n - 1;j ++)

61                 if(DP[i].x > DP[j].x && DP[i].y > DP[j].y && box < DP[j].z)

62                     box = DP[j].z;

63             DP[i].z += box;

64             max = max > DP[i].z ? max : DP[i].z;

65         }

66         printf("Case %d: maximum height = %d
",count,max);

67     }

68 

69     return    0;

70 }

71 

72 int    comp(const void * a,const void * b)

73 {

74     return    -(((struct x *)a) -> x - ((struct x *)b) -> x);

75 }