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. 注解:@Autowired

    Spring的bean对象自动装配注解,@Autowired有三种使用方式,下面分别介绍! 1. 作用于类成员变量上,即在定义类的成员变量的时候,至于其上方. public class Mobile{ ...

  2. LINQ to SQL语句

    http://kb.cnblogs.com/page/42477/2/ 本系列文章导航 LINQ to SQL语句(1)之Where LINQ to SQL语句(2)之Select/Distinct ...

  3. 用JS获取DropDownList选中得值

    HTML: <asp:DropDownList ID="DropdownList1" runat="server" AutoPostBack=" ...

  4. SSH_框架整合2—查询显示

    4. 完成功能. (1)com.atguigu.ssh.actions包下新建EmployeeAction.java package com.atguigu.ssh.actions; import j ...

  5. (转)jQuery禁止右键菜单,全选

    本文转载自:http://www.cnblogs.com/lucker/archive/2012/09/21/2696464.html $("body").bind("c ...

  6. NeHe OpenGL教程 第二十六课:反射

    转自[翻译]NeHe OpenGL 教程 前言 声明,此 NeHe OpenGL教程系列文章由51博客yarin翻译(2010-08-19),本博客为转载并稍加整理与修改.对NeHe的OpenGL管线 ...

  7. 设置session的生存时间

    1.最有效的方法 修改php.ini中  #session的超时时间 session.gc_maxlifetime = 1440  2.通过php去修改这个值 $Lifetime = 3600; $S ...

  8. dos查看端口

    C:\Documents and Settings\Administrator>netstat -an | findstr "22"  TCP    192.168.16.2 ...

  9. Win8下修改任務欄的資源管理器默認打開位置

    不能像win7一樣右鍵屬性改了,但還是有辦法的. 新建一個文件夾,建立快捷方式,右鍵快捷方式,將目標改為%windir%\explorer.exe /n,/e,D:\Desktop 然後將該快捷方式拖 ...

  10. JAVA 继承 extends

    /* 继承 1.提高了代码的复用性,简化了代码 2.让类与类之间产生了继承关系,才有了后面的多态的特性的存在 注意:千万不要为了获取其它类的功能简化代码,而建立继承关系, 必须要类与类之间存在继承关系 ...