UVA

The Tower of Babylon

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    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 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

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**因为刚刚开始学动态规划，这道题目也是花了我挺长时间的。
因为输入的是一个立方体，而立方体又有长宽高，所以要转化成矩形套矩形的问题的话，要先枚举立方体不同的三个面。然后根据紫书上的公式写了一个dp用的函数。最后选取数组之中最大的一个数字就是我们要的答案了。**

#include <iostream>#include <cstring>using namespace std;struct node{    int x,y;    int h;}v[110];int n,k,dp[110];bool pd(node a,node b){    if((a.x>b.x&&a.y>b.y)||(a.x>b.y&&a.y>b.x)) return true;    else return false;}int ddp(int r){    if(dp[r]!=-1) return dp[r];    dp[r]=v[r].h;    for(int i=0;i<k;i++)        if( pd(v[i],v[r]) ) dp[r] = max(dp[r], ddp(i) + v[r].h);    return dp[r];}int main(){    int Case=1;    while(cin>>n&&n)    {        int t=n,maxx=0;        k=0;        while(t--)        {            int a,b,c;            cin>>a>>b>>c;            v[k++]={ a, b, c };            v[k++]={ a, c, b };            v[k++]={ c, b, a };        }        memset(dp,-1,sizeof(dp));        for(int i=0;i<k;i++)            maxx= max(maxx,ddp(i));        cout<<"Case "<<Case++<<": maximum height = "<<maxx<<endl;    }    return 0;}