HDU-1069-Monkey and Banana

E - Monkey and Banana
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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Status

Practice

HDU 1069
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

题意：多组数据输入，第一行输入n，接着n行给出n组坐标，x，y，z，分别是长方体的长宽高，这三个数字的顺序可以颠倒，随便一个数都可以作为长方体的长或宽或高，而且这种型号的长方体有无数个。(以为无数多个就可以无数往上摞？）
把长方体摞一块，要求上面长方体的长和宽都要小于下面长方体的长和宽，输出能够摞出的最大高度。

思路：接收到x，y，z这三个数字后，组合出三种长方形，把新的长方体的数据存起来，然后就是简单的动规；

状态转移方程：dp[i]=max(dp[i],dp[j]+num[i].h)
如果看不懂的话，可以接着看下面的代码
代码

#include<iostream>#include<algorithm>#include<math.h>#include<string>#include<string.h>#include<stdio.h>#include<queue>using namespace std;//动规int dp[1005];//既是动规数组，也用来临时接收坐标struct node{    int x;//长    int y;//宽    int h;//高度} num[1005]; //存储坐标bool cmp(node a,node b){    if(a.x>b.x) return true;    if(a.x==b.x&&a.y>b.y) return true;    return false;}int main(){    int N;//N组坐标    int count=1;    while(~scanf("%d",&N)&&N)    {        memset(num,0,sizeof(num));        int length=0;//记录实际产生坐标数量        for(int i=0; i<N; i++)        {            scanf("%d%d%d",&dp[0],&dp[1],&dp[2]);            sort(dp,dp+3);            num[length].x=dp[2];            num[length].y=dp[1];            num[length].h=dp[0];            length++;            num[length].x=dp[2];            num[length].y=dp[0];            num[length].h=dp[1];            length++;            num[length].x=dp[1];            num[length].y=dp[0];            num[length].h=dp[2];            length++;        }        sort(num,num+length,cmp);//三级排序        for(int i=0; i<length; i++)            dp[i]=num[i].h;//初始化dp数组        for(int i=length-2; i>=0; i--)        {            for(int j=i+1; j<length; j++)            {                if(num[i].x>num[j].x&&num[i].y>num[j].y)                {                    if(dp[i]<dp[j]+num[i].h)                        dp[i]=dp[j]+num[i].h;                }            }        }        int max_high=dp[0];        for(int i=0; i<length; i++)            if(dp[i]>max_high)                max_high=dp[i];        printf("Case %d: maximum height = %d\n",count++,max_high);    }}