ZOJ - 1093 Monkey and Banana

Monkey and Banana

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

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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

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

简单的dp题。

分析：因为每块积木最多有3个不同的底面和高度，因此先把每块积木看成三种不同的积木，每种积木只有一个底面和一个高度。n种类型的积木转化为3*n个不同的积木的叠加，对这3 * n个积木的长边从大到小排序；接下来的问题就是找到一个递减的子序列，使得子序列的高度和最大即可。


数组dp：dp[i]表示是以第i块积木为顶的塔的最大高度


因此可得状态转移方程：dp[i] = max(dp[i]，dp[j] + r[i].z)（满足积木j的底面长和宽都大于积木i的长和宽）

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n;
struct R
{
    int x,y,z;
}r[100];
int dp[100];
bool cmp(const R a,const R b)
{
    if(a.x==b.x) return a.y>b.y;
    else return a.x>b.x;
}
int DP()
{
    int maxh=0,i,j;
    sort(r,r+n,cmp);
    for(i=0;i<n;i++)
    {
        dp[i]=r[i].z;
        for(j=i-1;j>=0;j--)
            if(r[j].x>r[i].x&&r[j].y>r[i].y)
            if(dp[j]+r[i].z>dp[i])
            dp[i]= dp[j]+r[i].z;
        maxh=maxh>dp[i]?maxh:dp[i];
    }
    return maxh;
}


int main()
{
    int i,j,a,b,c,kase=0;
    while(scanf("%d",&n)&&n!=0)
    {
        for(i=0,j=0;i<n;i++)
        {
           scanf("%d%d%d",&a,&b,&c);
           r[j].x=min(a,b);
           r[j].y=max(a,b);
           r[j].z=c;
           r[j+1].x=min(a,c);
           r[j+1].y=max(a,c);
           r[j+1].z=b;
           r[j+2].x=min(b,c);
           r[j+2].y=max(b,c);
           r[j+2].z=a;
           j+=3;
        }
        n=j;
        printf("Case %d: maximum height = %d\n",++kase,DP());
    }
    return 0;
}