HDU 1069 Monkey and Banana

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10163    Accepted Submission(s): 5282

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

 

这个题目的难点就在于形式的转化吧。我们拿到的是一种形态的箱子。对于这种箱子，他是可以摆出不同的长宽高的。比如样例给出的10 20 30，其中可以摆出长为30，宽为20，高为10；长为30，宽为10，高为20以及长为20，宽为10，高为30（因为长一定是比宽要长的，所以只有三种情况）所以我们就可以将这个题所有的箱子的状态都存入数组中，然后排序。这样就可以转化成为求最长下降子序列的问题。

代码如下：

/*************************************************************************
	> File Name: Monkey_and_banana.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Thu 22 Oct 2015 11:15:53 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

int n;
struct Node{
    int a, b, c;
}blocks[100];

bool cmp(Node x, Node y){
    if(x.a > y.a)
        return true;
    else if(x.a == y.a && x.b > y.b)
        return true;
    else 
        return false;

}

int dp[100];

int main(void){
    int cas = 0;
    while(1){
        scanf("%d", &n);
        if(n == 0)
            break;
        cas ++;
        int ans = 0;
        int e[3];
        for(int i = 0; i < n; i ++){
            scanf("%d%d%d",&e[0], &e[1], &e[2]);
            sort(e, e + 3);
            blocks[ans].a = e[2];
            blocks[ans].b = e[1];
            blocks[ans].c = e[0];
            ans ++;
            blocks[ans].a = e[2];
            blocks[ans].b = e[0];
            blocks[ans].c = e[1];
            ans ++;
            blocks[ans].a = e[1];
            blocks[ans].b = e[0];
            blocks[ans].c = e[2];
            ans ++;
        }
        int Max = 0;
        sort(blocks, blocks + ans, cmp);
        for(int i = 0; i < ans; i ++)
            dp[i] = blocks[i].c;
        for(int i = ans - 2; i >= 0; i --){
            for(int j = i + 1; j <= ans - 1; j ++){
                if(blocks[i].a > blocks[j].a && blocks[i].b > blocks[j].b){
                    dp[i] = dp[i] > dp[j] + blocks[i].c ? dp[i] : blocks[i].c + dp[j];
                }
            }
        }
        for(int i = 0; i < ans; i ++)
            Max = Max > dp[i] ? Max : dp[i];
        printf("Case %d: maximum height = %d\n",cas, Max);
    }
    return 0;
}