ACM HDU 1069 Monkey and Banana (动态规划）

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2599    Accepted Submission(s): 1345

Problem 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

 

Source

University of Ulm Local Contest 1996

 

Recommend

JGShining

 

 

/*
HDU1069Monkey and Banana
题目：给出一些长方体，然后让你把他堆成塔，
要求下面的塔的要比上面的塔大（长和宽），
而且每一种长方体的数量都是无限的。

此题目考察到动态规划里的最长有序子序列，

*/
#include<stdio.h>
#include<algorithm>
const int MAXN=200;
using namespace std;
struct Block
{
    int x,y,high;
    int dp;//该箱子在最下面时的最大高度 
}b[MAXN];

bool cmp(Block a,Block b)//用sort函数排序，先按x后按y升序
{
    if(a.x<b.x) return 1;
    else if(a.x==b.x&&a.y<b.y)  return 1;
    else return 0;
}        
int main()
{
    int n,i,x,y,z,j,k;
    int iCase=0;
    while(scanf("%d",&n),n)
    {
        iCase++;
        k=0;
        while(n--)
        {
            scanf("%d%d%d",&x,&y,&z);
            //把给出的block放置的所有可能放进block[]中，这样就可以解决有无限块的问题
            if(x==y)
            {
                if(y==z)//三个相等，放一个就够了
                {
                    b[k].x=x;b[k].y=y;b[k].high=z;b[k].dp=b[k].high;k++;
                }
                else  //x==y!=z时三种放法
                {
                    b[k].x=x;b[k].y=y;b[k].high=z;b[k].dp=b[k].high;k++;
                    b[k].x=z;b[k].y=y;b[k].high=x;b[k].dp=b[k].high;k++;
                    b[k].x=y;b[k].y=z;b[k].high=x;b[k].dp=b[k].high;k++;
                }                         
            }
            else
            {
                if(y==z)//三种放法
                {
                    b[k].x=x;b[k].y=y;b[k].high=z;b[k].dp=b[k].high;k++;
                    b[k].x=y;b[k].y=x;b[k].high=z;b[k].dp=b[k].high;k++;
                    b[k].x=y;b[k].y=z;b[k].high=x;b[k].dp=b[k].high;k++;
                } 
                else if(x==z)
                {
                    b[k].x=x;b[k].y=y;b[k].high=z;b[k].dp=b[k].high;k++;
                    b[k].x=y;b[k].y=x;b[k].high=z;b[k].dp=b[k].high;k++;
                    b[k].x=x;b[k].y=z;b[k].high=y;b[k].dp=b[k].high;k++;
                }  
                else//三个不等6种放法 
                {
                    b[k].x=x;b[k].y=y;b[k].high=z;b[k].dp=b[k].high;k++;
                    b[k].x=y;b[k].y=x;b[k].high=z;b[k].dp=b[k].high;k++;
                    b[k].x=x;b[k].y=z;b[k].high=y;b[k].dp=b[k].high;k++;
                    b[k].x=z;b[k].y=x;b[k].high=y;b[k].dp=b[k].high;k++;
                    b[k].x=y;b[k].y=z;b[k].high=x;b[k].dp=b[k].high;k++;
                    b[k].x=z;b[k].y=y;b[k].high=x;b[k].dp=b[k].high;k++;
                }          
            }        
        } 
        sort(b,b+k,cmp);
        int maxh=0;
        for(i=1;i<k;i++)
        {
            for(j=0;j<i;j++)
               if(b[i].x>b[j].x&&b[i].y>b[j].y)
                  b[i].dp=max(b[j].dp+b[i].high,b[i].dp);
            if(b[i].dp>maxh)maxh=b[i].dp;
        } 
        printf("Case %d: maximum height = %d\n",iCase,maxh);      
    }  
    return 0;  
}