Monkey and Banana

Monkey and Banana

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Time Limit: 2 Seconds Memory Limit: 65536 KB

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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 (x_i, y_i, z_i). 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 Specification

The input file will contain one or more test cases. The first line of each test case contains an integern,
representing the number of different blocks in the following data set. The maximum value forn is 30.
Each of the next n lines contains three integers representing the valuesx_i,y_i and z_i.
Input is terminated by a value of zero (0) for n.

Output Specification

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 "Casecase: maximum height =height"

Sample Input

?

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

1

1020 30

2

68 10

55 5

7

11 1

22 2

33 3

44 4

55 5

66 6

77 7

5

3141 59

2653 58

9793 23

8462 64

3383 27

0

Sample Output

?

1

2

3

4

Case1: maximum height = 40

Case2: maximum height = 21

Case3: maximum height = 28

Case4: maximum height = 342

#include<string>#include<climits>using namespace std; int box[100][3];int height[100];int number;</climits></string></iostream></pre><pre class="brush:java;">//转换砖块为三种类型void change(int index,int a,int b,int c){     box[index][0] = a;     box[index][1] = b;     box[index][2] = c;}</pre><pre class="brush:java;">//dpint DP(int j){    int& ans = height[j];    //搜索过直接返回    if(ans != -1)return ans;    </pre><pre class="brush:java;">   //搜索每一层    for(int i = 1; i <= number; i++)    {            if(i!=j)            {                 if((box[i][0] < box[j][0] && box[i][1] < box[j][1]) ||                  (box[i][1] < box[j][0] && box[i][0] < box[j][1]))                 {                    int  temp = DP(i) + box[i][2];                    if(temp > ans) ans = temp;                        }            }    }</pre><pre class="brush:java;">    //未更新，说明是最后一层。    if(ans == -1) ans = 0;    return ans;}int main(){         int n;    int Case = 0;    while(cin >> n && n)    {                 int a,b,c;        number = 0;    //地板看成无穷大砖块。        box[0][0] = box[0][1] = box[0][2] = INT_MAX;             for(int i = 1; i <= n; i++)        {                cin >> a >> b >> c;                change(++number,a,b,c);                change(++number,b,c,a);                change(++number,c,a,b);        }                 for(int i = 0; i <= number; i++)        {                height[i] = -1;        }        cout << "Case " << ++Case << ": maximum height = " << DP(0) << endl;    }    return 0;}