POJ 2241 The Tower of Babylon

The Tower of Babylon

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on  PKU. Original ID: 2241
64-bit integer IO format: %lld      Java class name: Main

Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contest, we will tell you the whole story: 
The babylonians had 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 wanted to construct the tallest tower possible by stacking blocks. The problem was 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. 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 babylonians can build with a given set of blocks.

 

Input

The input 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

Source

Ulm Local 1996

 

解题：动态规划，很像LIS

 1 #include <bits/stdc++.h>

 2 using namespace std;

 3 const int maxn = 10010;

 4 struct Cube{

 5     int x,y,z;

 6     Cube(int a = 0,int b = 0,int c = 0){

 7         x = a;

 8         y = b;

 9         z = c;

10     }

11     bool operator<(const Cube &t) const{

12         return x*y > t.x*t.y;

13     }

14 }cube[maxn];

15 int dp[maxn];

16 int main(){

17     int n,tot,x,y,z,kase = 1;

18     while(scanf("%d",&n),n){

19         for(int i = tot = 0; i < n; ++i){

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

21             cube[tot++] = Cube(x,y,z);

22             cube[tot++] = Cube(y,z,x);

23             cube[tot++] = Cube(x,z,y);

24         }

25         memset(dp,0,sizeof dp);

26         sort(cube,cube+tot);

27         int maxHeight = 0;

28         for(int i = 0; i < tot; ++i){

29             dp[i] = cube[i].z;

30             for(int j = i-1; j >= 0; --j){

31                 if(cube[j].x > cube[i].x && cube[j].y > cube[i].y)

32                     dp[i] = max(dp[i],dp[j] + cube[i].z);

33                 if(cube[j].y > cube[i].x && cube[j].x > cube[i].y)

34                     dp[i] = max(dp[i],dp[j] + cube[i].z);

35             }

36             maxHeight = max(maxHeight,dp[i]);

37         }

38         printf("Case %d: maximum height = %d\n",kase++,maxHeight);

39     }

40     return 0;

41 }

View Code