HDU——Monkey and Banana 动态规划

Monkey and Banana

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

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 (x_i, y_i, z_i). 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 Specification

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 x_i, y_i and z_i.
Input is terminated by a value of zero (0) for n.

Output Specification

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

Sample Output

Case 1: maximum height = 40

Case 2: maximum height = 21

Case 3: maximum height = 28

Case 4: maximum height = 342

动态规划第一题！

本题一开始我都没看懂哪里该用动态规划。而且关于砖块这里的处理，我也很伤脑筋。

解题思路是这样的：

每一种砖块都当成三种（底面积分别是长宽，长高，宽高）。。存入结构体数组中

然后按底面积大小进行排序。

动规的过程是这样的：

循环 1 到 n*3个砖块

再嵌套一个循环获取当前砖块下的，最大高度

加到砖块高度上

这样，遍历完成后，就会存贮了最大高度

#include <iostream>

#include <cstdio>

#include <algorithm>

using namespace std;

struct node

{

    int x,y,z,h;

}block[];

bool cmp2(int a,int b)

{

    return a<b;

}

bool cmp(node a,node b)

{

    return a.x*a.y<b.x*b.y;

}

int main()

{

    int n;

    int count=;

    while (scanf("%d",&n)!=EOF&&n)

    {

        int a[];

        int cur=;

        for (int i=;i<n;i++)

        {

            scanf("%d %d %d",&a[],&a[],&a[]);  //这里对长宽高也要排一下顺序，升序。

            sort(a,a+,cmp2);

            block[cur].x=a[];block[cur].y=a[];block[cur++].z=a[];

            block[cur].x=a[];block[cur].y=a[];block[cur++].z=a[];

            block[cur].x=a[];block[cur].y=a[];block[cur++].z=a[];

        }

        sort(block,block+cur,cmp);

        block[].h=block[].z;

        int max=;

        for (int j=;j<cur;j++) //遍历所有的砖块

        {

            max=;

            for (int k=;k<j;k++)

            {

                if (block[k].h>max&&block[k].x<block[j].x&&block[k].y<block[j].y)//寻找该砖块下的最大高度值

                 max=block[k].h;             //max当中保存的即为前j-1个砖块的最大高度值。

            }

            block[j].h=block[j].z+max;//找到当前最高值，加入到当前砖块的h中。

        }

        max=;

        for (int q=;q<cur;q++)

        {

            if (max<block[q].h) max=block[q].h;

        }

        count++;

        printf("Case %d: maximum height = %d\n",count,max);

    }

    return ;

}