UVa 437 --- The Tower of Babylon

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 blo

ck 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 "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

分析 : DAG最长路算法 

状态总数O(n) 每个状态决策O（n)  时间复杂度O（n^2) 

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#define max(a,b) (a>b?a:b)#define REP(i,n) for(int i = 0; i < (n) ; ++i)const int maxn = 30 + 10 ;int ra[maxn][3] , ans[maxn][maxn] ;             // ra[i][k] i：长方形序号 ， k： 高的序号using namespace std;int n ;void get_rac( int* racn , int n , int k ) {     // 引用指针    int num = 0 ;    REP( i , 3) {    if(  k != i ) racn[num++] = ra[n][i] ;      // 找到长宽 ， 并存储    }}int dp( int i , int j ) {    int& Lmax =  ans[i][j] ;                    // 引用ans[i][j]    if( Lmax > 0 ) return Lmax ;                // 记忆化搜索    Lmax = 0 ;    int rac1[3] , rac2[3] ;    get_rac( rac1 , i  , j ) ;    REP( ii , n ) {        REP( jj , 3) {            get_rac(rac2, ii , jj) ;            if( rac1[0] < rac2[0] && rac1[1] < rac2[1] )       // 严格小于                Lmax = max( Lmax, dp( ii, jj )+ra[ii][jj] ) ;  // 这样写  ， 整个函数返回值，不包含根节点 ， 在主函数加上        }    }   //也可写成Lmax= max (Lmax , dp(ii, jj)） ；循环外 Lmax += ra[i][j] ; return Lmax ;        // 因为 int& Lmax = ans[i][j]  ， 故不能写成 return Lmax + ra[i][j] ，ans[i][j]没改变 ;        // 这样写，函数递归完成包含根节点 ，主函数不必再加    return Lmax  ;}int main() {    int text = 1 ;    while( ~scanf("%d",&n),n) {        memset(ans,0, sizeof(ans)) ;        REP(i,n) {            REP(j,3){             scanf("%d",&ra[i][j] ) ;   // 存取三边分别作为高            }sort( ra[i], ra[i] + 3) ;  // 边从小到大排序        }        int maxx = 0  ;        REP(i, n ) {            REP( j , 3 ) {               maxx =   max(dp( i,j)+ra[i][j], maxx)  ;            }        }        printf("Case %d: maximum height = %d\n",text++, maxx) ;    }    return 0 ;}