HDU1069 最长单调子序列变形

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1907    Accepted Submission(s): 986

Problem 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

 

题目要求是 , 把给定尺寸的长方体 ( 数量不限 ) 叠在一起 , 能叠加的条件是 , 上面一个长方体的长宽比下面一个长方体的长宽短都短 , 或者一边相等 , 另一边较短 , 长方体可以任意面朝下摆放 , 求这些长方体能够叠成的最高的高度 .

不难看出 , 这题和最长递减子序列很相似 . 可以用类似的思路求解 .

首先对各种尺寸的长方体不同面朝下的各种摆法按照地面的长宽和面积排序 , 排序后 , 保证排在前面的长方体的底面的长和宽都比下一个的要长 , 或者一边相等 , 另一边较长 , 并且前面的长方体的面积必定比后面的大 .

设 h[i] 为前 i 个长方体中 , 包含第 i 个长方体的能够摆放的最大高度 ,z[i] 为第 i 个长方体的高 , 其转移方程为 :

h[i]=max(h[k])+z[i](1<=k<i, 且第 i 个长方体能摆在第 k 个长方体之上 )

代码如下 :

#include<stdio.h> #include<string.h> #include<algorithm> using std::sort; #define MAXN 100 struct S { int x,y,z; bool operator < (const S a) const { if ((x<a.x&&y<a.y)||(x<a.y&&y<a.x)||(x*y<a.x*a.y)) return 0; return 1; } }s[MAXN]; bool small (S a,S b) { if ((a.x<b.x&&a.y<b.y)||(a.x<b.y&&a.y<b.x)) return 1; return 0; } int main() { int cas,t,i,j,x,y,z,n,h[MAXN],max,maxx,maxi,maxj; int pre[MAXN]; cas=0; while (scanf("%d",&n)&&n) { ++cas; t=0; memset(pre,0,sizeof(pre)); for (i=1;i<=n;++i) { scanf("%d%d%d",&x,&y,&z); s[++t].x=x;s[t].y=y;s[t].z=z; s[++t].x=y;s[t].y=z;s[t].z=x; s[++t].x=x;s[t].y=z;s[t].z=y; } sort(s+1,s+t+1); maxx=0; for (i=1;i<=t;++i) {h[i]=s[i].z;if (h[i]>maxx) maxx=h[i];} for (i=1;i<=t;++i) { max=0; for (j=1;j<i;++j) { if (s[i].z+h[j]>max&&small(s[i],s[j])) max=s[i].z+h[j]; } if (max>h[i]) h[i]=max; if (max>maxx) maxx=max; } printf("Case %d: maximum height = %d/n",cas,maxx); } return 0; }