HDU 1069 Monkey and Banana （动态规划+LIS）

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

110 20 3026 8 105 5 571 1 12 2 23 3 34 4 45 5 56 6 67 7 7531 41 5926 53 5897 93 2384 62 6433 83 270

Sample Output

Case 1: maximum height = 40Case 2: maximum height = 21Case 3: maximum height = 28Case 4: maximum height = 342

 【题解】这道题和我前几天做的那个题原理一样，今天又重温了一下，感觉更加巩固了。

  题意是猴子要拿到一个奖品（具体啥东西没看懂），又够不着，现在给它几种垫脚石，都是长方体，每种垫脚石个数不限，可以垒起来，但前提是两块石头接触面要严格满足下接触面的长和宽大于（ps：没有等于）上接触面的长和宽，（给猴子留出踩脚的地方），现在要从这n种石头中选出一些垫脚石，使其可以累加得到最大高度，输出最大高度。

  【AC代码】

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[200],m;int xx,yy,zz;struct tree{    int x,y,z;    void input(int a,int b,int c)    {        x=a;        y=b;        z=c;    }}node[200];int max(int x,int y){    return x>y?x:y;}bool cmp(const tree a,const tree b){    return a.x*a.y<b.x*b.y;}int cnt;int main(){    int t=0;    while(~scanf("%d",&m),m)    {        cnt=0;        for(int i=1;i<=m;i++)        {            scanf("%d%d%d",&xx,&yy,&zz);            node[cnt++].input(xx,yy,zz);            node[cnt++].input(xx,zz,yy);            node[cnt++].input(yy,xx,zz);            node[cnt++].input(yy,zz,xx);            node[cnt++].input(zz,xx,yy);            node[cnt++].input(zz,yy,xx);        }        sort(node,node+cnt,cmp);        int ans=0;        dp[0]=0;        for(int i=0;i<cnt;i++)        {            dp[i]=node[i].z;            for(int j=0;j<i;j++)            {                if(node[i].x>node[j].x&&node[i].y>node[j].y)                    dp[i]=max(dp[i],dp[j]+node[i].z);//这儿可以理解为拿着一块砖，放在不同高度的墙上，找到其中最高的高度值            }                                        //dp[j]就是高度不一样的墙 ，node[i].z就是那块砖            if(ans<dp[i])                ans=dp[i];        }        printf("Case %d: maximum height = %d\n",++t,ans);    }    return 0;}

用了象形比喻，希望可以帮助理解。