POJ 2241 The Tower of Babylon（DP）

The Tower of Babylon

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2552		Accepted: 1425

Description

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

【思路】

一道经典的动归题，值得一做。开始想到了背包，虽然木块数是无限的，但是规定下面的木块横向的边长得大于上面的木块，且由于每个木块不会超过三种摆法，最多只能做到同时采用两个一样规格的木块，所以猜测也许是个多重背包。如果把三种摆法单独拿出来，每一种摆法视作一个木块，应该就可以化作01背包了。而问题在于，木块的叠放显然是有顺序关联的，想要分阶段去决策，就得利用好木块之间存在的偏序某种关系，那就是底面的长和宽。做法是，把木块按照底面积排序，先叠底小的，再往下放底大的，dp[i]保存第i个木块作为底层木块塔的高度，直接更新即可。看到网上有人用记忆化搜索的思路做，感觉还是太盲目了，不如先排序，再分阶段决策省事。

【代码】

#include
#include
#include
using namespace std;

const int MAXN=100,INF=0x3f3f3f3f;

struct block{
    int x,y,z;

    bool operator<(const block &another)const{
        return x*yb[j].x&&b[i].y>b[j].y||b[i].x>b[j].y&&b[i].y>b[j].x)
                    dp[i]=max(dp[i],dp[j]+b[i].z);
            ans=max(ans,dp[i]);
        }
        printf("Case %d: maximum height = %d\n",++kase,ans);
    }
    return 0;
}