hdu 1069 Monkey and Banana

一天刷三道动规小练，这月刚好可以把动规小练AK，当然晚上有时候还会有比赛就不能刷题了，所以要加油啊，这道题还好吧，不算太难，老是控制不住看别人博客的习惯，看一道题没思路，就老是想看别人的博客，唉~~不过这道题还好，自己想了几种方法，结果总能发现自己想的方法的欠缺的地方，最后终于自己想出来了，题意是给你几个长方体（每一种无限个），问如果叠加起来，最多能叠加多高，下面的长方体的底面积的长和宽要严格大于上面的长方体的长和宽，这样的话，每一个长方体都有三种状态，即高不一样，然后分别按照长，宽按从小到大排序，然后把所有长方体遍历一遍，对应每一个长方体，从目前位置到最小的长方体再遍历一遍，求出长和宽分别小于自己的长方体的最大高，然后给对应的长方体加上这个高，这样跑一遍，最后在从所有的长方体里再求最大的那个高，然后就是所求的高，（不知道能不能大一时参加省赛，唉~~~~）

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11658    Accepted Submission(s): 6085

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
   
   
   
   
   
   
   
   


   
   
   
   

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define swap(a,b) a^=b^=a^=b
#define max(a,b) a>b?a:b
using namespace std;
struct stu
{
    int x,y,h;
} a[100];
int cmp(stu a,stu b)
{
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;
}
int main()
{
    int n,S=0;
    while(scanf("%d",&n)&&n)
    {
        int i,k=0,j,aa,bb,cc;
        for(i=1; i<=n; i++)
        {
            scanf("%d%d%d",&aa,&bb,&cc);
            a[++k].x=aa,a[k].y=cc,a[k].h=bb;
            a[++k].x=cc,a[k].y=bb,a[k].h=aa;
            a[++k].x=bb,a[k].y=aa,a[k].h=cc;
        }
        for(i=1; i<=k; i++)
            if(a[i].x<a[i].y)
                swap(a[i].x,a[i].y);
        sort(a+1,a+k+1,cmp);
        for(i=2; i<=k; i++)
        {
         int d=0;
            for(j=i-1; j>=1; j--)
                if(a[j].x<a[i].x&&a[j].y<a[i].y&&a[j].h>d)
                {
                 d=a[j].h;
                }
                a[i].h+=d;
        }
        int  maxh=0;
        for(i=1; i<=k; i++)
        {
            if(a[i].h>maxh)
                maxh=a[i].h;
        }
        printf("Case %d: maximum height = ",++S);
        printf("%d\n",maxh);
    }
}