Monkey and Banana

最新推荐文章于 2025-08-10 14:30:24 发布

转载最新推荐文章于 2025-08-10 14:30:24 发布 · 55 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Tinamei/p/4731839.html

文章标签：

#人工智能

本文介绍了一道关于通过不同尺寸的方块堆叠以达到特定高度的问题，并使用动态规划（DP）来解决这一问题。该问题源于一项研究猴子智商的实验，通过允许猴子将方块堆叠起来拿到香蕉，挑战在于确定能够堆叠的最大高度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit 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

10 20 30

6 8 10

5 5 5

1 1 1

2 2 2

3 3 3

4 4 4

5 5 5

6 6 6

7 7 7

31 41 59

26 53 58

97 93 23

84 62 64

33 83 27

Sample Output

Case 1: maximum height = 40

Case 2: maximum height = 21

Case 3: maximum height = 28

Case 4: maximum height = 342

题意：不同石头摞起来求最大高度，dp（摞的时候下边的长和宽要比上边的大，dp【i】存的是i块石头放在最下边的时候，总的石头摞起来所能到达的最大高度

 1 ///#pragma comment (linker, "/STACK:102400000,102400000")
 2 
 3 #include <iostream>
 4 #include <queue>
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <algorithm>
 8 #include <cmath>
 9 #include <cstdlib>
10 #include <limits>
11 #include <stack>
12 #include <vector>
13 #include <map>
14 
15 using namespace std;
16 
17 #define N 1350
18 #define INF 0xfffffff
19 #define PI acos (-1.0)
20 #define EPS 1e-8
21 
22 struct node
23 {
24     int l, w, h, v;
25 }P[N];
26 
27 bool cmp(node a, node b)
28 {
29     return a.v < b.v;
30 }
31 
32 int main()
33 {
34     int n, i, x, y, z, t = 1;
35     int dp[N];
36 
37     while(scanf("%d", &n), n)
38     {
39         i = 0;
40         memset(dp, 0, sizeof(dp));
41 
42         while(n--)
43         {
44             scanf("%d%d%d", &x, &y, &z);
45             P[i].l = x, P[i].w = y, P[i].h = z, P[i++].v = x * y;
46             P[i].l = y, P[i].w = z, P[i].h = x, P[i++].v = z * y;
47             P[i].l = z, P[i].w = x, P[i].h = y, P[i++].v = x * z;
48         }                       // 说是石头无限利用，其实顶多用3次，因为摞的时候有要求
49         sort(P, P+i, cmp);
50         int ans = 0;
51         for(int j = 0; j < i; j++)
52         {
53             dp[j] = P[j].h;
54             for(int k = 0; k < j; k++)  // j只能放在比它小的石头下边
55             {
56                 if(P[j].l > P[k].l && P[j].w > P[k].w || P[j].l > P[k].w && P[j].w > P[k].l)   // 满足摞的条件，可以放下边，dp【i】存的是第i块石头放最下边的时候最大高度
57                     dp[j] = max(dp[j], dp[k]+P[j].h);  // 如果j可以放k下边，那么dp【j】更新
58             }
59             ans = max(ans, dp[j]);   // 求哪块石头放在下边是最大值
60         }
61         printf("Case %d: maximum height = %d\n", t++, ans);
62     }
63     return 0;
64 }