时间限制:10000ms
单点时限:1000ms
内存限制:256MB

描述

Little Hi gets lost in the city. He does not know where he is. He does not know which direction is north.

Fortunately, Little Hi has a map of the city. The map can be considered as a grid of N*M blocks. Each block is numbered by a pair of integers. The block at the north-west corner is (1, 1) and the one at the south-east corner is (N, M). Each block is represented by a character, describing the construction on that block: '.' for empty area, 'P' for parks, 'H' for houses, 'S' for streets, 'M' for malls, 'G' for government buildings, 'T' for trees and etc.

Given the blocks of 3*3 area that surrounding Little Hi(Little Hi is at the middle block of the 3*3 area), please find out the position of him. Note that Little Hi is disoriented, the upper side of the surrounding area may be actually north side, south side, east side or west side.

输入

Line 1: two integers, N and M(3 <= N, M <= 200).
Line 2~N+1: each line contains M characters, describing the city's map. The characters can only be 'A'-'Z' or '.'.
Line N+2~N+4: each line 3 characters, describing the area surrounding Little Hi.

输出

Line 1~K: each line contains 2 integers X and Y, indicating that block (X, Y) may be Little Hi's position. If there are multiple possible blocks, output them from north to south, west to east.

样例输入
8 8
...HSH..
...HSM..
...HST..
...HSPP.
PPGHSPPT
PPSSSSSS
..MMSHHH
..MMSH..
SSS
SHG
SH.
样例输出
5 4

代码实现:

 #include <stdio.h>
#include <stdlib.h>
#include <string.h>
int compare(char** s1, int i, int j, char surround[][])
{
for (int a = i; a< + i; a++)
for (int b = j; b< + j; b++)
{
if (s1[a][b] != surround[a - i][b - j])
return ;
} return ;
}
int main()
{
int n = , m = ;
scanf("%d %d", &n,&m);
getchar();
char** map = (char**)malloc(sizeof(char)*n);
for (int i = ; i<n; i++)
{
map[i] = (char*)malloc(sizeof(char)*m);
fgets(map[i], m+, stdin);
getchar();
}
char surround[][] = { };
for (int i = ; i<; i++)
{
fgets(surround[i], , stdin);
getchar();
}
char surround1[][] = { };
char surround2[][] = { };
char surround3[][] = { };
char position = surround[][];
for (int i = ; i<; i++)
{
for (int j = ; j<; j++)
{
surround1[ - j][i] = surround[i][j];
}
}
for (int i = ; i<; i++)
{
for (int j = ; j<; j++)
{
surround2[ - j][i] = surround1[i][j];
}
}
for (int i = ; i<; i++)
{
for (int j = ; j<; j++)
{
surround3[ - j][i] = surround2[i][j];
}
}
for (int i = ; i<n - ; i++)
{
for (int j = ; j<m - ; j++)
{ if (map[i][j] == position)
{
if (compare(map, i - , j - , surround))
{
printf("%d %d\n", i + , j + );
continue;
}
if (compare(map, i - , j - , surround1))
{
printf("%d %d\n", i + , j + );
continue;
}
if (compare(map, i - , j - , surround2))
{
printf("%d %d\n", i + , j + );
continue;
}
if (compare(map, i - , j - , surround3))
{
printf("%d %d\n", i + , j + );
continue;
}
}
}
}
return ;
}

问题:

使用Visual Studio编译运行样例结果正确,但是提交代码通过G++运行会出现Wrong Answer,若知道原因的话还望请留言告知!

#1094 : Lost in the City by C solution的更多相关文章

  1. hihoCoder #1094 : Lost in the City(枚举,微软苏州校招笔试 12月27日 )

    #1094 : Lost in the City 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 Little Hi gets lost in the city. He ...

  2. hihocoder #1094 : Lost in the City微软苏州校招笔试 12月27日 (建图不大【暴力枚举】 子图的4种形态 1Y )

    #1094 : Lost in the City 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 Little Hi gets lost in the city. He ...

  3. #1094 : Lost in the City

    时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 Little Hi gets lost in the city. He does not know where he is ...

  4. hihoCoder#1094 Lost in the City

    原题地址 限时10s,所以不用考虑什么算法了,暴力吧 分别按照3x3视野的四个方向去地图上匹配,把符合的地点标记出来,最后统一按照从上到下,从左到右的顺序输出. 代码: #include <io ...

  5. 【BZOJ2001】 [Hnoi2010]City 城市建设

    BZOJ2001 [Hnoi2010]City 城市建设 Solution 我们考虑一下这个东西怎么求解? 思考无果...... 咦? 好像可以离线cdq,每一次判断一下如果这条边如果不选就直接删除, ...

  6. WATERHAMMER: A COMPLEX PHENOMENON WITH A SIMPLE SOLUTION

    开启阅读模式 WATERHAMMER A COMPLEX PHENOMENON WITH A SIMPLE SOLUTION Waterhammer is an impact load that is ...

  7. bzoj:1687;poj 2434:[Usaco2005 Open]Navigating the City 城市交通

    Description A dip in the milk market has forced the cows to move to the city. The only employment av ...

  8. [Swift]LeetCode807. 保持城市天际线 | Max Increase to Keep City Skyline

    In a 2 dimensional array grid, each value grid[i][j]represents the height of a building located ther ...

  9. [Swift]LeetCode1029. 两地调度 | Two City Scheduling

    There are 2N people a company is planning to interview. The cost of flying the i-th person to city A ...

随机推荐

  1. 06jQuery-04-DOM操作

    jQuery既然是为了帮助你能从js的繁琐中解脱出来,自然在DOM操作上也有自己的一套. 1.修改Text和HTML 之前我们提到过,如果用JS的话,你要修改Text或者HTML需要用到其innerH ...

  2. markdown编辑器的学习

    markdown编辑器的学习 1 标题 一级标题 二级标题 三级标题 四级标题 五级标题 六级标题 2列表 无序列表 1 2 3 4 有序列表 1 2 3 4 3引用 这里是引用,哈哈我也不知道到我引 ...

  3. Java并发之线程间的协作

    上篇文章我们介绍了synchronized关键字,使用它可以有效的解决我们多线程所带来的一些常见问题.例如:竞态条件,内存可见性等.并且,我们也说明了该关键字主要是一个加锁和释放锁的集成,所有为能获得 ...

  4. linux中文乱码

    txt文件在linux环境下打开呈现了乱码状态. 解决方法1:在linux用iconv命令,如乱码文件名为zhongwen.txt,那么在终端输入如下命令: iconv -f gbk -t utf8 ...

  5. (转)添加PROPAGATION_REQUIRES_NEW 事务没有产生作用

    最近在做事务添加时  发现自己的事务没有新建,上网查到   仅用作收藏. 其二  注意  事务的注解  应该在 内层的事务上面 一.描述 Spring遇到嵌套事务时,当被嵌套的事务被定义为" ...

  6. 这是一名Java学者关于学习方向的建议

    无可厚非,编程是一门艺术.但是辉煌的背后必须是一段辛苦的奋斗过程,而过程的引导方向就是最重要的一环.Java语言可谓是引领了编程的潮流,你会是怎样去学的呢? 这是一名Java学者的学习方向的建议 注: ...

  7. SQL Server 2016 Alwayson新增功能

    标签:SQL SERVER/MSSQL SERVER/数据库/DBA/ 概述 SQLServer2016发布版本到现在已有一年多的时间了,目前最新的稳定版本是SP1版本.接下来就开看看2016在Alw ...

  8. 【转】String字符串相加的问题

    String字符串相加的问题 前几天同事跟我说我之前写的代码中在操作字符串时候,使用字符串相加的方式而不是使用StringBuffer或者StringBuilder导致内存开销很大.这个问题一直在困扰 ...

  9. Theano学习-scan循环

    \(1.Scan\) 通用的一般形式,可用于循环 减少和映射(对维数循环)是特殊的 \(scan\) 对输入序列进行 \(scan\) 操作,每一步都能得到一个输出 \(scan\) 能看到定义函数的 ...

  10. JavaWeb(四)EL表达式

    前言 前面详细的说明了什么是JSP和它的一些元素,这篇给大家介绍一下的是EL表达式. 用EL表达式,能更好的使用JSP中的各种内置对象和作用域. 楼主作为大四狗马上要出去面试了,内心很紧张!!! 一. ...