HDU 1069 Monkey and Banana （动态规划、上升子序列最大和）

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24147    Accepted Submission(s): 12938

 

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

题目大意

给出一些长方体的规格，每种长方体都有无限个，问利用这些长方体（长宽高可以转换），下层的长和宽都比上层的大，最多能堆多高

题目分析

虽说是无限多个，但是由于长和高堆起来都不能相等，所以一个规格的长方体最多能用6个（长宽高的不同排列），这样就将题目转化成了一个有限的木块的题目。

首先将木块按照x值与y值从大到小排序，用dp[i]来记录i为最顶端木块时，所能达到的最大高度。

由于我们已经将木块排好序了，那么处理到 i 木块时，能在这个木块底下的就只有 [0, i-1] 这些木块，所以用一个 j 从0循环到 i-1：

如果i的x和y小于j的x和y，那么dp[i]显然等于：max ( dp[j] + a[i].h , dp[i] )

求出dp[i]时，要注意更新结果。

代码

#include <bits/stdc++.h>  

using namespace std; 

typedef struct 
{
    int x;
    int y;
    int h;
}node;
node a[205];

int i,n,num,x,y,z,t=0,anss,dp[200],j;

bool cmp (node a,node b)
{
    if(a.x>b.x)
    return 1;
    else if(a.x==b.x&&a.y>b.y)
    return 1;
    else return 0;
}

int main()
{
    while(scanf("%d",&n),n!=0)
    {
        t++;
        memset(a,0,sizeof(a));
        num=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d %d %d",&x,&y,&z);
            //cout<<x<<y<<z;
            a[num].x=x,a[num].y=y,a[num].h=z;
            num++;
            a[num].x=y,a[num].y=x,a[num].h=z;
            num++;
            a[num].x=z,a[num].y=x,a[num].h=y;
            num++;
            a[num].x=z,a[num].y=y,a[num].h=x;
            num++;
            a[num].x=x,a[num].y=z,a[num].h=y;
            num++;
            a[num].x=y,a[num].y=z,a[num].h=x;
            num++;
        }
        sort(a,a+num,cmp);
        anss=0;
        memset(dp,0,sizeof(dp));
        for(i=0;i<num;i++)
        dp[i]=a[i].h;
        for(i=0;i<num;i++)
        {
            for(j=0;j<i;j++)
            {
                if(a[i].x<a[j].x&&a[i].y<a[j].y)
                dp[i]=max(dp[j]+a[i].h,dp[i]);
            }
            if(dp[i]>anss)
            anss=dp[i];
        }
        printf("Case %d: maximum height = %d\n",t,anss); 
    }
}