Problem C

Character Recognition?

Write a program that recognizes characters. Don't worry, because you only need to recognize three digits: 1, 2 and 3. Here they are:

.*.  ***  ***
.*. ..* ..*
.*. *** ***
.*. *.. ..*
.*. *** ***

Input

The input contains only one test case, consisting of 6 lines. The first line contains n, the number of characters to recognize (1<=n<=10). Each of the next 5 lines contains 4n characters. Each character contains exactly 5 rows and 3 columns of characters followed by an empty column (filled with '.').

Output

The output should contain exactly one line, the recognized digits in one line.

Sample Input

3
.*..***.***.
.*....*...*.
.*..***.***.
.*..*.....*.
.*..***.***.

Output for the Sample Input

123

The Ninth Hunan Collegiate Programming Contest (2013) Problemsetter: Rujia Liu Special Thanks: Feng Chen, Md. Mahbubul Hasan

 方法很多,深入理解dfs即可方便解决此题 ,我觉得这个方法比较好,有一定的价值。

#include <iostream>
#include <stdio.h>
#include <queue>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <algorithm>
#include <map>
#include <stack>
#include <math.h>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std ;
typedef long long LL ; struct Point{
int X ;
int Y ;
Point(){} ;
Point(int x ,int y):X(x),Y(y){} ;
}; char str[][] ;
int N ;
int d[][]={{,},{-,},{,-},{,}} ;
bool visited[][] ;
vector<Point>my_hash[] ; int cango(int x ,int y){
return <=x&&x<=&&<=y&&y<=*N&&str[x][y]=='*';
} void dfs(int x ,int y ,int color){
my_hash[color].push_back(Point(x,y)) ;
visited[x][y]= ;
for(int i=;i<;i++){
int xx=x+d[i][] ;
int yy=y+d[i][] ;
if(cango(xx,yy)&&!visited[xx][yy]){
dfs(xx,yy,color) ;
}
}
} int main(){
int color= ;
scanf("%d",&N) ;
for(int i=;i<=;i++)
scanf("%s",str[i]+) ;
for(int i=;i<=N;i++)
my_hash[i].clear() ;
for(int i=;i<=;i++)
for(int j=;j<=*N;j++){
if(str[i][j]=='*'&&!visited[i][j]){
color++ ;
dfs(i,j,color) ;
}
} for(int i=;i<=N;i++){
Point first = my_hash[i][] ;
Point second = my_hash[i][my_hash[i].size()-] ;
if(first.Y==second.Y&&first.X+==second.X)
putchar('') ;
else if(first.Y<second.Y)
putchar('') ;
else if(first.X+==second.X)
putchar('') ;
}
puts("") ;
return ;
}

The Ninth Hunan Collegiate Programming Contest (2013) Problem C的更多相关文章

  1. The Ninth Hunan Collegiate Programming Contest (2013) Problem A

    Problem A Almost Palindrome Given a line of text, find the longest almost-palindrome substring. A st ...

  2. The Ninth Hunan Collegiate Programming Contest (2013) Problem F

    Problem F Funny Car Racing There is a funny car racing in a city with n junctions and m directed roa ...

  3. The Ninth Hunan Collegiate Programming Contest (2013) Problem H

    Problem H High bridge, low bridge Q: There are one high bridge and one low bridge across the river. ...

  4. The Ninth Hunan Collegiate Programming Contest (2013) Problem I

    Problem I Interesting Calculator There is an interesting calculator. It has 3 rows of button. Row 1: ...

  5. The Ninth Hunan Collegiate Programming Contest (2013) Problem J

    Problem J Joking with Fermat's Last Theorem Fermat's Last Theorem: no three positive integers a, b, ...

  6. The Ninth Hunan Collegiate Programming Contest (2013) Problem G

    Problem G Good Teacher I want to be a good teacher, so at least I need to remember all the student n ...

  7. The Ninth Hunan Collegiate Programming Contest (2013) Problem L

    Problem L Last Blood In many programming contests, special prizes are given to teams who solved a pa ...

  8. German Collegiate Programming Contest 2013:E

    数值计算: 这种积分的计算方法很好,学习一下! 代码: #include <iostream> #include <cmath> using namespace std; ; ...

  9. German Collegiate Programming Contest 2013:B

    一个离散化的简单题: 我用的是STL来做的离散化: 好久没写离散化了,纪念一下! 代码: #include<cstdio> #include<cstring> #include ...

随机推荐

  1. Android启动画面实现

    每个Android应用程序启动之后都会出现一个Splash启动界面,显示产品LOGO.公司LOGO或者开发者信息.如果应用程序启动时间比较长,那么启动界面就是一个很好的东西,可以让用户耐心等待这段枯燥 ...

  2. 解决每次打开office 2010显示正在配置的问题

    解决每次打开office 2010显示正在配置的问题 OFFICE 2010 MSDN版出来后,下载安装 ,启动后发现每次打开都会出现“正在配置”的进度,删除重装亦不成功,对SETUP.EXE改名也不 ...

  3. 【转】Dubbo是Alibaba开源的分布式服务框架

    Dubbo是Alibaba开源的分布式服务框架,它最大的特点是按照分层的方式来架构,使用这种方式可以使各个层之间解耦合(或者最大限度地松耦合).从服务模型的角度来看,Dubbo采用的是一种非常简单的模 ...

  4. Perl system(cmd) 和 `cmd` 的区别探讨

    在perl中系统调用有两种方式,一种是system(cmd),另一种是`system`以前一直没注意,这两种方式的区别,还以为是一样的,今天写脚本的时候,忽然想要获取命令的返回值,然后,用了my $r ...

  5. Oracle-decode函数

    decode函数 简单例子:管理员登录Oracle select sid, username, decode(command, 0, 'None', 2, 'Insert', 3, 'Select', ...

  6. 关于Linux中exec的一点心得

    最近在学习linux操作系统中的相关知识,在使用execlp系统调用时,发现了些有趣的东西. 首先,关于execlp函数的用法: int execlp(const char *file, const ...

  7. (C/C++ interview) Static 详解

    C Static http://stackoverflow.com/questions/572547/what-does-static-mean-in-a-c-program Static could ...

  8. Scala第二章学习笔记

    最基本的练习~: 使用伴生对象: object holder{ class Foo{ private var x = 5} object Foo{def im_in_yr_foo(f: Foo) = ...

  9. IREP_SOA Integration SOAP概述(概念)

    20150827 Created By BaoXinjian

  10. python(22)总结下最近遇到的编码问题

    最近爬取,或者解析网页是总是遇到编码问题(我的版本:python2.7) 一.常见异常:UnicodeEncodeError: 'ascii' codec can't encode character ...