Monkey and Banana hdu（1069）

Monkey and Banana

TimeLimit: 2000/1000 MS(Java/Others)    MemoryLimit: 65536/32768 K (Java/Others)
Total Submission(s):3594    AcceptedSubmission(s): 1843

Problem Description

A group of researchers are designing an experiment to test the IQof a monkey. They will hang a banana at the roof of a building, andat the mean time, provide the monkey with some blocks. If themonkey is clever enough, it shall be able to reach the banana byplacing one block on the top another to build a tower and climb upto get its favorite food.

The researchers have n types of blocks, and an unlimited supply ofblocks of each type. Each type-i block was a rectangular solid withlinear dimensions (xi, yi, zi). A block could be reoriented so thatany two of its three dimensions determined the dimensions of thebase and the other dimension was theheight. 

They want to make sure that the tallest tower possible by stackingblocks can reach the roof. The problem is that, in building atower, one block could only be placed on top of another block aslong as the two base dimensions of the upper block were bothstrictly smaller than the corresponding base dimensions of thelower block because there has to be some space for the monkey tostep on. This meant, for example, that blocks oriented to haveequal-sized bases couldn't be stacked. 

Your job is to write a program that determines the height of thetallest tower the monkey can build with a given set ofblocks.

 

Input

The input file will contain one or more test cases. The first lineof each test case contains an integer n,
representing the number of different blocks in the following dataset. The maximum value for n is 30.
Each of the next n lines contains three integers representing thevalues 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 (theyare numbered sequentially starting from 1) and the height of thetallest possible tower in the format "Case case: maximum height =height".

 

Sample Input

1 

10 2030 

2

6 810 

5 55 

7

1 11 

2 22 

3 33 

4 44 

5 55 

6 66 

7 77 

5 

31 4159 

26 5358 

97 9323 

84 6264 

33 8327 

0

 

Sample Output

Case1: maximum height = 40 

Case2: maximum height = 21 

Case3: maximum height = 28 

Case4: maximum height = 342

 

Source

University of Ulm Local Contest 1996

#include <stdio.h>#include<algorithm>#include<iostream>#define MAX 1000using namespace std;int max(int a,int b){    return a>b?a:b;}struct ar{    int a,b,h,s;}area[3000];int cmp(ar s,ar t){    return s.s>t.s;}int x[MAX],y[MAX],z[MAX],dp[MAX];int main(){    int n,i,j ,w = 1;    while(~scanf("%d",&n)&&n){        int h=0;        for(i=0;i<n;i++)            scanf("%d%d%d",&x[i],&y[i],&z[i]);        for(i=0;i<n;i++)            for(j=0;j<3;j++){                area[h].a = x[i];                area[h].b = y[i];                area[h].h = z[i];                area[h].s = x[i]*y[i];                h++;                swap(x[i],z[i]);                swap(x[i],y[i]);            }        //for(i=0;i<h;i++)        //  printf("%d %d %d%d\n",area[i].a,area[i].b,area[i].h,area[i].s);        sort(area,area+h,cmp);        int ans=0;        dp[0]=area[0].h;        for(i=1;i<h;i++){            dp[i]=area[i].h;            for(j=0;j<i;j++){                if(( area[i].a < area[j].a &&area[i].b < area[j].b )||( area[i].a < area[j].b&&area[i].b<area[j].a))                    dp[i]=max(dp[i],dp[j]+area[i].h);                            }            if(dp[i]>ans)                ans=dp[i];        }        printf("Case %d: maximum height =%d\n",w++,ans);    }    return 0;}