[UVa 437]The Tower of Babylon

Problem 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 tex2html_wrap_inline32 . 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 and Output

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 tex2html_wrap_inline40 , tex2html_wrap_inline42 and tex2html_wrap_inline44 .
Input is terminated by a value of zero (0) for n.
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
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

My Problem Report

经典DAG记忆化搜索模版题。叠放关系转化为二元关系，以此建图。在表示状态是需要采取一些技巧，《入门经典》里有细讲，这里不再累述。
这道题建立了一个非常经典的模型。与之前所做的记忆化搜索不同的是，这道题没有确定的起始状态和结束状态，所以只有采用递归的方法进行状态的转移，而不能采用递推（即使递推也非常麻烦）。

My Source Code

//  Created by Chlerry in 2015.
//  Copyright (c) 2015 Chlerry. All rights reserved.
//

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cstring>
#include <climits>
#include <string>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
using namespace std;
#define Size 100000
#define ll long long
#define mk make_pair
#define pb push_back
#define mem(array, x) memset(array,x,sizeof(array))
typedef pair<int,int> P;

vector<int> vx,vy,h;
int f[100],n;
int DFS(int u)
{
    if(f[u]) return f[u];
    else f[u]=h[u];
    for(int i=0;i<n;i++)
        if( (vx[u]<vx[i] && vy[u]<vy[i]) || (vx[u]<vy[i] && vy[u]<vx[i]) )
            f[u]=max(f[u],DFS(i)+h[u]);
    return f[u];
}
int main()
{
    freopen("in.txt","r",stdin);
    int x,y,z;
for(int ca=1;cin>>n && n;ca++)
{
    vx.clear(),vy.clear();h.clear();mem(f,0);
    for(int i=0;i<n;i++)
    {
        cin>>x>>y>>z;
        vx.pb(x);vy.pb(y);h.pb(z);
        vx.pb(x);vy.pb(z);h.pb(y);
        vx.pb(y);vy.pb(z);h.pb(x);
    }
    n*=3;
    for(int i=0;i<n;i++)
        DFS(i);
    printf("Case %d: maximum height = ",ca);
    cout<<*max_element(f,f+n)<<endl;
}
    return 0;
}