首页 > ACM题库 > HDU-杭电 > HDU 1069 Monkey and Banana-动态规划-[解题报告] C++
2013
11-26

HDU 1069 Monkey and Banana-动态规划-[解题报告] C++

Monkey and Banana

问题描述 :

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.

输入:

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.

输出:

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".

样例输入:

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

样例输出:

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

http://acm.hdu.edu.cn/showproblem.php?pid=1069

/*

题目大意:给出n种类型的木块,求用这些木块可以堆起的最大高度,要求上边的木快的长和宽都要严格小于下边。

*/

#include<stdio.h>
#include<stdlib.h>
#define N 95
int f[N];  //f[N]记录加上第i个木块后的最大高度
struct X
{
	int x,y,z;
}block[N];
int cmp(const struct X* a,const struct X* b)
{
	if((*a).x!=(*b).x)
		return (*a).x-(*b).x;
	else
		return (*a).y-(*b).y;
}
int main()
{
	int T,n,a,b,c,i,j,temp,tallest;
	T=1;
	while(scanf("%d",&n),n){
		for(i=0,j=0;j<n;j++){       //每种木块可以有三种放法
			scanf("%d%d%d",&a,&b,&c);
			block[i].x=a; block[i].y=b; block[i].z=c;
			block[i+1].x=a; block[i+1].y=c; block[i+1].z=b;
			block[i+2].x=c; block[i+2].y=b; block[i+2].z=a;
			i+=3;
		}
		for(i=0;i<n*3;i++){          //找出每种木块的长和宽
			if(block[i].x<block[i].y){
				temp=block[i].x;
				block[i].x=block[i].y;
				block[i].y=temp;
			}
		}
		qsort(block,n*3,sizeof(block[0]),cmp); //先按木块的"长"升序排列,"长"相等时再按"宽"升序排列
		for(i=0,tallest=0;i<3*n;i++){    //将f[i]初始化为第i个的高度,计算f[i]时,遍历0—>i,找出放上第i个木块时的最大高度。
			f[i]=block[i].z;
			for(j=0;j<=i;j++){
				if(block[i].x>block[j].x&&block[i].y>block[j].y){
					f[i]=max(f[i],f[j]+block[i].z);
				}
				tallest=max(tallest,f[i]);
			}
		}
		printf("Case %d: maximum height = %d\n",T++,tallest);
	}
	return 0;
}


  1. 题本身没错,但是HDOJ放题目的时候,前面有个题目解释了什么是XXX定律。
    这里直接放了这个题目,肯定没几个人明白是干啥

  2. bottes vernies blanches

    I appreciate the efforts you men and women place in to share blogs on such sort of matters, it was certainly useful. Keep Posting!