POJ 1101 译文
The Game
题意:
Description
One morning, you wake up and think: “I am such a good programmer. Why not make some money?” So you decide to write a computer game.
一天清晨,你起床后想到:“我是一个这么牛的程序员,为什么不赚点钱呢?”所以,你决定写一个电脑游戏。
The game takes place on a rectangular board consisting of w * h squares. Each square might or might not contain a game piece, as shown in the picture.
这个游戏是在一个长方形板上的,这个板由w*h个小正方形组成。每一个小正方形可能会包括一个游戏的块,如下图所示。
One important aspect of the game is whether two game pieces can be connected by a path which satisfies the two following properties:
这个游戏的一个最重要的方面是判断两个块会不会由符合下列要求的路径连接。
It consists of straight segments, each one being either horizontal or vertical.
1.路径是由直的部分组成,每个部分或水平或垂直。
It does not cross any other game pieces.
2.路径不能跨过其它的游戏方块。
(It is allowed that the path leaves the board temporarily.)
(路径可以暂时离开板)
Here is an example:
下面 举个例子。
The game pieces at (1,3) and at (4, 4) can be connected. The game pieces at (2, 3) and (3, 4) cannot be connected; each path would cross at least one other game piece.
(1,3)和(2,4)可以被连接。(2,3)和(3,4)不能被连接。
The part of the game you have to write now is the one testing whether two game pieces can be connected according to the rules above.
游戏的一部分 是你必须写一个程序证明两个木块是否能够被连接起来。
这里写图片描述
Input
输入
The input contains descriptions of several different game situations. The first line of each description contains two integers w and h (1 <= w,h <= 75), the width and the height of the board. The next h lines describe the contents of the board; each of these lines contains exactly w characters: a “X” if there is a game piece at this location, and a space if there is no game piece.
输入是多种情况的,第一行包括两个整数w和h(q<=w,h<=75),即板的宽度和长度,下h行表示板上都有什么。每行包括w个字符,’X’表示有一个游戏方块,空格表示没有游戏方块。
Each description is followed by several lines containing four integers x1, y1, x2, y2 each satisfying 1 <= x1,x2 <= w, 1 <= y1,y2 <= h. These are the coordinates of two game pieces. (The upper left corner has the coordinates (1, 1).) These two game pieces will always be different. The list of pairs of game pieces for a board will be terminated by a line containing “0 0 0 0”.
每种情况后包括几行询问,每行包括4个整数X1,Y1,X2,Y2(1<=X1,X2<=w;I<=Y1,Y2<=h)。这是2个游戏片段的坐标。(左上角为坐标(1,1)。) 这两个坐标将永远是不同的。“0 0 0 0”表示输入结束。
The entire input is terminated by a test case starting with w=h=0. This test case should not be procesed.
整个输入结束 是输入了(0,0) 这个case不用考虑。
Output
输出
For each board, output the line “Board #n:”, where n is the number of the board. Then, output one line for each pair of game pieces associated with the board description. Each of these lines has to start with “Pair m: “, where m is the number of the pair (starting the count with 1 for each board). Follow this by “ksegments.”, where k is the minimum number of segments for a path connecting the two game pieces, or “impossible.”, if it is not possible to connect the two game pieces as described above.
每个板,输出一行“Board #n:”,n是第n块板。然后输出一行为每对坐标是否相关联。如果关联,输出“Pair m: k segments.”m为这块板的第m次询问,k为路径包括几个部分。否则输出“Pair m: impossible.”
Output a blank line after each board.
每个板询问完后输出个空行。
Sample Input样例输入
5 4
XXXXX
X X
XXX X
XXX
2 3 5 3
1 3 4 4
2 3 3 4
0 0 0 0
0 0
Sample Output样例输出
Board #1:
Pair 1: 4 segments.
Pair 2: 3 segments.
Pair 3: impossible.
POJ 1101 译文的更多相关文章
- OpenJudge 2802 小游戏 / Poj 1101 The Game
1.链接地址: http://bailian.openjudge.cn/practice/2802 http://poj.org/problem?id=1101 2.题目: 总时间限制: 1000ms ...
- 搜索 + 剪枝 --- POJ 1101 : Sticks
Sticks Problem's Link: http://poj.org/problem?id=1011 Mean: http://poj.org/problem?id=1011&lan ...
- POJ 1101 简单BFS+题意
The Game 题意: Description One morning, you wake up and think: "I am such a good programmer. Why ...
- POJ - 1101 The Game dfs
题意:给你一个地图,上面有一些‘X',给你起点终点,让你输出从起点到终点的路径中转向(改变方向)次数最少的路径,注意,不能穿过别的’X'并且可以超过边界 题解:关于超过边界,只要在外围多加一圈‘ ’. ...
- POJ 1101
#include <iostream> #include <string> #define MAXN 78 #define min _min #define inf 12345 ...
- POJ 1101 The Game(BFS+判方向)
The Game Description One morning, you wake up and think: "I am such a good programmer. Why ...
- POJ 2570(floyd)
http://poj.org/problem?id=2570 题意:在海底有一些网络节点.每个节点之间都是通过光缆相连接的.不过这些光缆可能是不同公司的. 现在某个公司想从a点发送消息到b点,问哪个公 ...
- POJ 1426 Find The Multiple --- BFS || DFS
POJ 1426 Find The Multiple 题意:给定一个整数n,求n的一个倍数,要求这个倍数只含0和1 参考博客:点我 解法一:普通的BFS(用G++能过但C++会超时) 从小到大搜索直至 ...
- POJ 1006 Biorhythms (中国剩余定理)
在POJ上有译文(原文右上角),选择语言:简体中文 求解同余方程组:x=ai(mod mi) i=1~r, m1,m2,...,mr互质利用中国剩余定理令M=m1*m2*...*mr,Mi=M/mi因 ...
随机推荐
- css流光效果
css流光效果1: <!DOCTYPE html> <html> <head> <title>ww</title> </head> ...
- btrfs基础
btrfs文件系统简介 btrfs文件系统:技术预览版(Centos7) Btrfs(B-tree.Butter FS.Better FS),GPL授权,Orale2007提出是想用来取代Ext文 ...
- SqlServer大数据的分区方案
这里介绍的是大数据量的维护日志的分区解决方案: 每个月1张数据表,1个分组文件.31个分区(按每天1个分区).... 为了日后维护方便,直接删除旧的日志数据文件就可以,而不需要去做分区压缩. --用的 ...
- JSONP代码收藏
摘抄自jQuery,用于JSONP请求. var callback = 'callback_' + (new Date() - 0), url = 'http://localhost/', scrip ...
- Python 查看关键字
关键字 import keyword print(keyword.kwlist)
- Python - 模块(一)
目录 Python - 模块(一) 模块的引用方式 常用模块 random(随机模块) os模块 sys 序列化模块 hashlib subprocess optparse struct Python ...
- Haybale Guessing
Haybale Guessing Time Limit: 1000MS Memory Limit: 65536K Description The cows, who always ha ...
- hive计算日期差
首先,hive本身有一个UDF,名字是datediff.我们来看一下这个日期差计算的官方描述,(下面这个是怎么出来的): hive> desc function extended datedif ...
- hdu_hpu第八次周赛_1002 大菲波数_201310270958
大菲波数 Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total Submissi ...
- kendo Grid的toolbar自定义
由于这个toolbar官方进增加了create,save还有一个是_____ 所以想要自定义话就需要使用 下面的代码(这个是MVVM模式) data-toolbar='[{ template: Ken ...