ACM 动态规划 完全背包 E - Monkey and Banana

   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

1.看到这个题目的时候首先想到的是，这是完全背包的问题。

2.注意对长宽高进行排序，最后对面积也要排序。

3.注意，多次使用一个变量的时候，是否需要初始化的问题。比如这里的num，这一点一定要注意，错了不止一次了。

4.对于这个题目来说，一定要分析清楚，对于一个障碍物来说，就只有三种存在转化形态。（因为是最后要确定高度，所以x,y,z一定轮流做一次高。对于长度和宽度来说，交不交换一样，都是一种，故而总的来说就只有三种方式）

5.借鉴本题中num的使用，很巧妙。

6.总的思想来说就是先把所有的障碍物形态找到，然后再上升为背包问题。

#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
typedef struct bolck{
     int x;
     int y;
     int z;
     int are;
}block;


block b[303];
int a[3];
int dp[303];


int cmp(block a,block b)
{
    return a.are<b.are;


}
int main()
{
    int n,num;
    int flag=1;
    while(~scanf("%d",&n)&&n)
    {
        num=0;
        memset(dp,0,sizeof(dp));
        memset(b,0,sizeof(b));
        for(int i=0;i<n;i++)
        {


            scanf("%d%d%d",&a[0],&a[1],&a[2]);
            sort(a,a+3);
            b[num].x=a[0],b[num].y=a[1],b[num].z=a[2],b[num].are=b[num].x*b[num].y,num++;
            b[num].x=a[1],b[num].y=a[2],b[num].z=a[0],b[num].are=b[num].x*b[num].y,num++;
            b[num].x=a[0],b[num].y=a[2],b[num].z=a[1],b[num].are=b[num].x*b[num].y,num++;
        }
        sort(b,b+num,cmp);
        int nmax=0;
        for(int i=0;i<num;i++)
        {
            dp[i]=b[i].z;
            for(int j=0;j<i;j++)
            {


                if(b[j].x<b[i].x&&b[j].y<b[i].y)
                {
                    dp[i]=max(dp[i],dp[j]+b[i].z);


                }
            }
           nmax=max(dp[i],nmax);
        }
        printf("Case %d: maximum height = %d\n",flag++,nmax);
    }
    return 0;
}