Monkey and Banana 题解（动态规划）

Monkey and Banana 简单的动态规划

1.注：

　　本人第一篇博客，有啥不足还请多多包涵，有好的建议请指出。你以为有人读你博客，还给你提意见。

2.原题

　　Background：

　　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（为了好看，前面多了tab，复制时注意一下，第一次写博客，也不知影不影响复制）

　　Sample Output

　　Case 1: maximum height = 40 　　

　　Case 2: maximum height = 21 　　

　　Case 3: maximum height = 28 　　

　　Case 4: maximum height = 342

3.题目大意

　　猴子（为了还原背景。。。）有n种长方体（每种都没有数量限制）,长宽高都给你，让你帮猴子计算他能搭多高

　　限制：下面的长方体的长和宽都要大于上面的，当然哪条边是长，哪条边是宽，哪条边是高你说了算。

4.分析题目

　　猴子很聪明，他们先把问题简单化，我们也试试：

　　　　如果长宽高是给定的，我怎么解决这个问题?

　　友好城市这题做过吗？没有的话就先去做一下。类似的，我们这一题也可以像友好城市那样做。

　　要求长宽都是递增的，那就先对长或宽进行排序，然后再找另一个的最长上升子序列（当然，这里每个的权值就不一定是1了），找完之后，答案不就出来了吗？

　　猴子表示想到这里就可以了，但是我们是人类，要展现人类的智慧：

　　　　我们要把长方体进行旋转。

　　咋旋转呢，我们这样想，a，b，c三个，别管怎么放置，每种放置只用一次（因为要求严格上升吗），那就瞬间好办了：

　　　　我把一个长方体变成6个！！！

　　　　那不会用重了吗？哦，兄弟，非常抱歉，请你再读一遍题。

　　　　于是问题迎刃而解。

5.别整没用的，上代码

 #include <cstdio>
 #include <string>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 const int maxn=*+;
 int Dp[maxn];
 struct K{
     int x;
     int y;
     int h;
     friend bool operator < (K a,K b){
         return a.x<b.x;
     }
 }tin[maxn];//物品的结构体
 int main(){
     int n;
     int jsjs=;
     while(){//这样写大家习惯吗？
         jsjs++;
         scanf("%d",&n);
         if(!n)
             break;
         memset(Dp,,sizeof(Dp));
         for(int i=;i<=n;i++){
             scanf("%d%d%d",&tin[i].x,&tin[i].y,&tin[i].h);
             tin[i+n].x=tin[i].y;//这里可要想清楚，还有，可整个函数
             tin[i+n].y=tin[i].x;
             tin[i+n].h=tin[i].h;
             tin[i+n*].x=tin[i].y;
             tin[i+n*].y=tin[i].h;
             tin[i+n*].h=tin[i].x;
             tin[i+n*].x=tin[i].h;
             tin[i+n*].y=tin[i].y;
             tin[i+n*].h=tin[i].x;
             tin[i+n*].x=tin[i].x;
             tin[i+n*].y=tin[i].h;
             tin[i+n*].h=tin[i].y;
             tin[i+n*].x=tin[i].h;
             tin[i+n*].y=tin[i].x;
             tin[i+n*].h=tin[i].y;
         }
         sort(tin+,tin++*n);
         int ans=;
         for(int i=;i<=*n;i++){
             Dp[i]=tin[i].h;
             for(int j=;j<i;j++)//数据范围真的友好
                 if(tin[i].x>tin[j].x&&tin[i].y>tin[j].y)
                     Dp[i]=max(Dp[i],Dp[j]+tin[i].h);
             ans=max(ans,Dp[i]);
         }
         printf("Case %d: maximum height = %d\n",jsjs,ans);
     }
     return ;
 }