标签：printf mit com 保留 which init eve 实验 ble
STL是指C++的标准模板库。（存储着一些常用的算法和容器）
vector是一个不定长数组。它把一些常用的操作”封装“在vector类型内部。
例如，a是一个vector。1对元素的操作有，可以用a.size()读取它的大小，a.resize()改变它的大小，a.push_back()向尾部添加元素，a.pop_back()删除最后一个元素。2对数组的操作有：a.clear()清空，a.empty()测试是否为空。
vectors是一个模板类。
它的使用声明：vetor<int>a或者vector<double>b。
vector<int>a类似一个 int a[]的整型数组，而vector<string>类似一个 string a[]的字符串数组。
vector可以直接赋值，还可以作为函数的参数或者返回值。
例题5-2
Many areas of Computer Science use simple, abstract domains for both analytical and empirical studies. For example, an early AI study of planning and robotics (STRIPS) used a block world in which a robot arm performed tasks involving the manipulation of blocks.
在计算机科学中的很多地方都会使用简单，抽象的方法来做分析和实验验究。比如在早期的规划学和机器人学的人工智能研究就利用一个积木世界，让机械臂执行操作积木的任务。
In this problem you will model a simple block world under certain rules and constraints. Rather than determine how to achieve a specified state, you will "program" a robotic arm to respond to a limited set of commands.
在这个问题中，你将在确定的规则和约束条件下构建一个简单的积木世界。这不是让你来研究怎样达到某种状态，而是编写一个“机械臂程序”来响应有限的命令集。
The problem is to parse a series of commands that instruct a robot arm in how to manipulate blocks that lie on a flat table. Initially there are n blocks on the table (numbered from 0 to n-1) with block bi adjacent to block bi + 1 for all 0 ≤ i < n - 1 as shown in the diagram below:
问题就是分析一系列的命令，告诉机械臂如何操纵放在一个平台上的积木。最初平台上有n个积木（编号由0到n - 1），对于任意的0 ≤ i < n - 1，积木bi都与bi + 1相临，图示如下：
The valid commands for the robot arm that manipulates blocks are:
机械臂操作积木的有效指令列举如下：
Any command in which a = b or in which a and b are in the same stack of blocks is an illegal command. All illegal commands should be ignored and should have no affect on the configuration of blocks.
当a = b或a和b处在同一摞时，任何企图操作a和b的命令都是非法的。所有非法的命令都要忽略，且不能对当前积木的状态产生作用。
The input begins with an integer n on a line by itself representing the number of blocks in the block world. You may assume that 0 < n < 25.
输入由1个整数n开始开始，该整数独占一行，表示积木世界中的积木数量。你可以假定0 < n < 25。
The number of blocks is followed by a sequence of block commands, one command per line. Your program should process all commands until the quit command is encountered.
从积木数量值的下一行开始是一系列的命令，每条命令独占一行。你的程序要处理所有的命令直到输入退出命令。
You may assume that all commands will be of the form specified above. There will be no syntactically incorrect commands.
你可以假定所有的命令都按上文所示的格式给出。不会出现语法错误的命令。
The output should consist of the final state of the blocks world. Each original block position numbered i (0 ≤ i < n where n is the number of blocks) should appear followed immediately by a colon. If there is at least a block on it, the colon must be followed by one space, followed by a list of blocks that appear stacked in that position with each block number separated from other block numbers by a space. Don‘t put any trailing spaces on a line.
以积木世界的最终状态作为输出。每一个原始积木的位置i（0 ≤ i < n，n为积木数量）后面都要紧跟一个冒号。如果至少有一个积木在该位置上，冒号后面都要紧跟一个空格，然后是该位置上所有积木编号的序列。每2个积木的编号之间以一个空格隔开。行尾不能出现多余的空格。
There should be one line of output for each block position (i.e., n lines of output where n is the integer on the first line of input).
每个积木位置独占一行（即第一行输入的n，对应输出n行数据）。
10
move 9 onto 1
move 8 over 1
move 7 over 1
move 6 over 1
pile 8 over 6
pile 8 over 5
move 2 over 1
move 4 over 9
quit
0: 0
1: 1 9 2 4
2:
3: 3
4:
5: 5 8 7 6
6:
7:
8:
9:
每个木块堆的高度不确定，所以用vector来保存很合适；而木块堆的个数不超过n，所以用一个数组来存就可以了。
1 #include<cstdio> 2 #include<string> 3 #include<vector> 4 #include<iostream> 5 using namespace std; 6 7 const int maxn = 30; 8 int n; 9 vector<int>pile[maxn]; //每一个pile[i]是一个vector 10 11 12 //找木块a所在的pile和hight，以引用的形式返回调用者 13 void find_block(int a, int &p, int &h) 14 { 15 for ( p = 0; p < n; p++) 16 for (h = 0; h < pile[p].size(); h++) 17 if (pile[p][h] == a) 18 return; 19 } 20 21 //把第p堆高度为h的木块上方的所有木块移回原位 22 void clear_above(int p, int h) 23 { 24 for (int i = h + 1; i < pile[p].size(); i++) 25 { 26 int b = pile[p][i]; 27 pile[b].push_back(b); //把木块b放回原位 28 } 29 pile[p].resize(h + 1); //pile只应保留下标0~h的元素 30 } 31 32 //把第p堆高度为h及其上方的木块整体移动到p2堆的顶部 33 void pile_onto(int p, int h, int p2) 34 { 35 for (int i = h; i < pile[p].size(); i++) 36 pile[p2].push_back(pile[p][i]); 37 pile[p].resize(h); 38 } 39 40 void printf() 41 { 42 for (int i = 0; i < n; i++) 43 { 44 printf("%d:", i); 45 for (int j = 0; j < pile[i].size(); j++) 46 { 47 printf(" %d", pile[i][j]); 48 } 49 printf("\n"); 50 } 51 } 52 53 54 int main() 55 { 56 int a, b; 57 cin >> n; 58 string s1, s2; 59 for (int i = 0; i < n; i++) 60 pile[i].push_back(i); 61 while (cin >> s1) 62 { 63 if (s1 == "quit") 64 break; 65 cin >> a >> s2 >> b; 66 int pa, pb, ha, hb; 67 find_block(a, pa, ha); 68 find_block(b, pb, hb); 69 if (pa == pb) continue; //非法指令 70 if (s2 == "onto") clear_above(pb, hb); 71 if (s1 == "move") clear_above(pa, ha); 72 pile_onto(pa, ha, pb); 73 } 74 printf(); 75 return 0; 76 }
标签：printf mit com 保留 which init eve 实验 ble
原文地址：http://www.cnblogs.com/Strugglinggirl/p/5993993.html
