#1094 : Lost in the City by C solution
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
- 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 ...
- 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 ...
- #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 ...
- 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 ...
- 【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, ...
- 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 ...
- 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 ...
- [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 ...
- [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
- Spring - Spring容器概念及其初始化过程
引言 工作4年多,做了3年的java,每个项目都用Spring,但对Spring一直都是知其然而不知其所以然.鄙人深知Spring是一个高深的框架,正好近期脱离加班的苦逼状态,遂决定从Spring的官 ...
- Oracle单引号的用法-转义
在ORACLE中,单引号有两个作用: 1:字符串是由单引号引用 2:转义. 单引号的使用是就近配对.而在单引号充当转义角色时相对不好理解 1.从第二个单引号开始被视为转义符,如果第二个单引号后面还 ...
- Eclipse插件springsource-tool-suite在线和离线安装步骤
springsource-tool-suite插件是一个基于Eclipse的开发环境,为开发Spring应用程序而定制.它提供了一个即用的环境来实现,调试,运行和部署Spring应用程序,包括Pivo ...
- Opengl4.5 中文手册—B
索引 A B C D E F G H I J K L M N O P Q ...
- mongoDB学习手记2--建库、删库、插入、更新
上一篇 讲了在windows系统下的安装和启动,本文主要讲怎么建库.删库.插入.更新 在讲之前我们说一下mongoDB的一些基本概念,我们对比关系型数据库能更直观的理解 SQL术语/概念 Mongo ...
- 【Revit API】梁构件支座检查算法
一.前言 应该是第二次写关于Revit API的博文了.虽然在BIM企业中工作,从事桌面BIM软件开发,但是我是不怎么喜欢写Revit API相关的代码.平时更多的是在写界面展示,架构 ...
- JavaWeb(二)会话管理之细说cookie与session
前言 前面花了几篇博客介绍了Servlet,讲的非常的详细.这一篇给大家介绍一下cookie和session. 一.会话概述 1.1.什么是会话? 会话可简单理解为:用户开一个浏览器,点击多个超链接, ...
- linux上搭建ftp
linux上搭建ftp 重要 解决如何搭建ftp 解决用户指定访问其根目录 解决访问ftp超时连接 解决ftp主动连接.被动连接的问题 1.安装ftp ...
- php Socket通信
<?php error_reporting(0); $host = "0.0.0.0"; $port = 1082; $maxUser = 10; set_time_limi ...
- IE9总是弹出“ICBC Anti-Phishing class” 加载项是否要启用还是不启用的提示
解决方法: 后来在通过查询,发现 IcbcDaemon.exe 进程是写在系统服务中的,我们可以在系统的服务管理工具中停止该服务: 1.单击开始,在搜索框中输入 services.msc ,按下回车键 ...