hdu 1069-Monkey And Banana（The Tower of Babylon）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1069

 

题目描述：

 

Description

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 (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 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

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

 

 

 

题目大意：

给你n块砖块，每个砖块可以横着、侧着、竖着摆放，这里，叫你把这些砖块搭成一个塔状，要求下面的长和宽要比上面的大，问你，所能搭成的塔的最大高度

 

题目分析：

为了减少内存空间，我们先把长宽高排一下序，规定每个砖块摆放的时候都有长>宽，所以，每种砖块也就只有三种摆放情况，然后我们再把所有的砖块按照长降序，宽降序的顺序排一下序。到达这一步，我们就可以用传统的动态规划的思想来做这一题了（详见代码中的注释）

AC代码：

 

#include
#include
#include
#include
#include
#include
using namespace std;

struct node
{
    int x,y,z;//代表砖块的长宽高
}block[1005];

bool cmp(node a,node b)//按照砖块的长从大到小，宽从大到小排序
{
     if(a.x>b.x)
        return true;
     else if(a.x==b.x&&a.y>b.y)
        return true;
     return false;
}

int dp[1005];
int n,num=1;

int main()
{
    while(1)
    {
        cin>>n;
        if(n==0)
            break;
        memset(dp,0,sizeof(dp));
        memset(block,0,sizeof(block));
        int count=0,d[3];
        while(n--)
        {
            cin>>d[0]>>d[1]>>d[2];//一个砖块有三个面，这里默认长比宽大
            sort(d,d+3);//拍一下序，就只有三种情况，否则有六种情况
            block[count].x=d[2];block[count].y=d[1];block[count++].z=d[0];
            block[count].x=d[2];block[count].y=d[0];block[count++].z=d[1];
            block[count].x=d[1];block[count].y=d[0];block[count++].z=d[2];
        }
        sort(block,block+count,cmp);//对砖块进行排序（大->小）
        for(int i=0;i=0;i--)
        {
            for(int j=i+1;jblock[j].x&&block[i].y>block[j].y)||(block[i].x>block[j].y&&block[i].y>block[j].x))
                    {
                        dp[i]=max(dp[i],dp[j]+block[i].z);
                    }
            }
            Max=max(Max,dp[i]);
        }
        printf("Case %d: maximum height = ",num++);
        cout<