HDU 1069 Monkey and Banana （dp）

题目链接

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

分析：

题目倒是不难！对于动态规划的题，我觉得最重要的就是找出所谓的递归方程吧！

这点最为重要！找到了递归方程，代码实现只是时间问题罢了！

这道题目有一些要点要抓住！

1，下一层的面积要严格大于上一层！我在这里错了几次！所谓严格大于，那就是下一层的长宽都大于下一层的长宽或宽长，等于都不行！

2,这道问题比较像01背包问题！我用的方法是拆分！即将一种block拆成3种！（一种block有三种摆放姿势，且一种block最多放这三种姿势）这样，先按照面积排序！然后再次动归！可以减少时间复杂度！

3,hdp[i]表示如果以第i块block为顶，最多可以达到的高度！

代码：

#include<iostream>
#include<stdio.h>
#include <algorithm>
using namespace std;

typedef struct Block
{
    int x,y,high,dp;
};

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

int max(int a,int b)
{
    return a>b?a:b;
}
Block b[1000];
int main()
{
    int x,y,z,n,i,j,k,Max,p=0;
    while(~scanf("%d",&n)&&n)
    {
        p++;
        k=0;
        while(n--)
        {
            scanf("%d%d%d",&x,&y,&z);
            if(x==y)
            {
                if(y==z)
                {
                    b[k].x=x;
                    b[k].y=x;
                    b[k].high=x;
                    b[k++].dp=x;
                }
                else
                {
                    b[k].x=b[k].y=x;
                    b[k].high=z;
                    b[k++].dp=z;
                    b[k].x=x;
                    b[k].y=z;
                    b[k].high=x;
                    b[k++].dp=x;
                    b[k].x=z;
                    b[k].y=x;
                    b[k].high=x;
                    b[k++].dp=x;
                }
            }
            else
            {
                if(x==z)
                {
                    b[k].x=b[k].y=x;
                    b[k].high=y;
                    b[k++].dp=y;
                    b[k].x=x;
                    b[k].y=y;
                    b[k].high=x;
                    b[k++].dp=x;
                    b[k].x=y;
                    b[k].y=x;
                    b[k].high=x;
                    b[k++].dp=x;
                }
                else if(y==z)
                {
                    b[k].x=b[k].y=y;
                    b[k].high=x;
                    b[k++].dp=x;
                    b[k].x=x;
                    b[k].y=y;
                    b[k].high=y;
                    b[k++].dp=y;
                    b[k].x=y;
                    b[k].y=x;
                    b[k].high=y;
                    b[k++].dp=y;
                }
                else
                {
                    b[k].x=x;
                    b[k].y=y;
                    b[k].high=z;
                    b[k++].dp=z;
                    b[k].x=x;
                    b[k].y=z;
                    b[k].high=y;
                    b[k++].dp=y;
                    b[k].x=y;
                    b[k].y=x;
                    b[k].high=z;
                    b[k++].dp=z;
                    b[k].x=y;
                    b[k].y=z;
                    b[k].high=x;
                    b[k++].dp=x;
                    b[k].x=z;
                    b[k].y=x;
                    b[k].high=y;
                    b[k++].dp=y;
                    b[k].x=z;
                    b[k].y=y;
                    b[k].high=x;
                    b[k++].dp=x;
                }
            }
            sort(b,b+k,cmp);
            Max=0;
            for(i=0; 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);
                Max=max(b[i].dp,Max);
            }
        }
        printf("Case %d: maximum height = %d\n",p,Max);
    }
    return 0;
}