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CATM - The Cats and the Mouse |
In a rectangular field of size n by m squares there is a mouse and two cats. The mouse is the first to make a move, then each of the cats makes a move, then again its the mouse's turn, and so on. In each move both the mouse and the cats can move exactly one square vertically or horizontally. If the mouse is standing at the edge of the field then in its next move it can jump off the field and is saved from the cats. If in the next move one of the cats moves to the field with the mouse then there is no escape for the mouse ... =(
You are to write a program which, knowing the initial positions of mouse and the two cats, will find out if there is any way for the mouse to escape from the cats, assuming of course that each cat will do its best to catch the mouse.
Input
In the first line of input two integers n and m are given, not exceeding 100, where n is the number of rows, and m - the number of columns. The second line contains a number k [k <= 10], which defines the number of test cases for the given field. In the next k lines the initial positions of the mouse and the cats are given. The position in the field is given by two numbers: the first is the number of the row, the second is the number of the column. The first two integers are the coordinates of the mouse, the next four integers are the coordinates of the cats.
Output
You must output k lines with answers for each test case. The answer is YES, if the mouse can escape or NO otherwise.
Example
Input: 5 3 3 2 2 1 1 3 3 2 3 1 3 5 2 3 2 1 2 4 3 Output: NO YES YES
Author: Filimonenkov D.O.
Added by: | Roman Sol |
Date: | 2006-05-04 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO NODEJS PERL6 VB.NET |
Resource: | ZCon 2007 |
hide comments
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2022-10-31 15:06:00
If You Are Stuck This Is Very Simple Solution Without BFS : <snip> [Simes]: No thanks Last edit: 2022-10-31 22:28:58 |
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2021-10-04 11:17:20
nice problem acc without bfs Last edit: 2021-10-04 11:18:30 |
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2020-09-09 09:54:34
excellent question, kudos to the maker |
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2020-07-15 01:16:26
AC in one go but i didn't use bfs. only conditions :)) |
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2020-06-11 08:56:11
This problem is intended to do with graph traversals like bfs or dfs and not to use the straight forward method of distance finding and checking the conditions |
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2020-06-08 10:44:38
accepted in 1 go ..shouldnt be on spoj ...dissapointed |
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2019-08-24 11:35:01
not includes bfs not at all |
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2019-04-20 14:47:32
got accepted with simple implementation.but how do i solve using bfs apart from applying bfs thrice for each of the cat and mouse? |
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2019-01-13 13:49:01
Only 4 comparisions are enough .... :) ... AC Iin one go :) |
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2018-12-21 21:13:22
AC in 1 GO :)) |