Monkey and Banana

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。

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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

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: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 }

 

转载于:https://www.cnblogs.com/Tinamei/p/4731839.html

### 实现猴子摘香蕉问题的 C 语言程序 为了实现猴子移动到箱子并最终拿到香蕉的过程,完整的解决方案应包括以下几个部分: 1. **定义数据结构** 使用结构体来表示房间内的各个对象及其属性。这有助于清晰地管理不同实体的状态。 2. **初始化状态** 设置初始条件,如猴子、箱子和香蕉的具体位置。 3. **目标检测** 定义何时认为任务已完成的标准,即当猴子站在箱顶且能触及香蕉时。 4. **动作执行** 编写一系列操作函数用于改变当前场景中的物体相对位置关系,比如`move_monkey()` 和 `push_box()` 5. **主循环控制** 创建一个主循环来进行交互式的命令解析与响应机制,直到达成目的为止。 以下是基于上述原则构建的一个简化版C语言源码框架[^1]: ```c #include <stdio.h> #include <stdbool.h> // Define the structure to hold positions of objects. typedef struct { int monkey; int box; int banana; } State; void move_monkey(State *s, int new_pos); bool can_reach_banana(const State* s); int main(void){ // Initialize state with given locations (example values). State initialState = { .monkey=0 , .box=2 , .banana=3 }; printf("Initial setup:\nMonkey at %d\nBox at %d\nBananas at %d.\n", initialState.monkey, initialState.box, initialState.banana ); while (!can_reach_banana(&initialState)){ char action[8]; scanf("%7s",action); // Read user input if(strcmp(action,"go")==0 || strcmp(action,"Go")==0){ int pos; scanf("%d",&pos); move_monkey(&initialState,pos); printf("Moved monkey to position:%d\n",initialState.monkey); } else{ puts("Invalid command."); } // Print updated status after each step printf("\nCurrent Status:\n"); printf("Monkey is now at %d\n", initialState.monkey); printf("The Box remains at %d\n", initialState.box); printf("And Bananas are still hanging from point %d\n", initialState.banana); } puts("Goal achieved! The monkey has reached the bananas!"); } /// Function implementations below... void move_monkey(State *s,int newPos){ (*s).monkey=newPos; } bool can_reach_banana(const State* s){ return ((*s).monkey==(*s).box && (*s).box==(*s).banana)?true:false ; } ``` 此代码实现了基本功能,允许通过简单的命令行界面指导猴子行动直至成功获取香蕉。然而,这段代码仅提供了一个基础版本;实际应用可能还需要考虑更多细节,例如更复杂的环境建模或者错误处理等特性。
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