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. maven本地仓库的配置以及如何修改默认.m2仓库位置

    本地仓库是远程仓库的一个缓冲和子集,当你构建Maven项目的时候,首先会从本地仓库查找资源,如果没有,那么Maven会从远程仓库下载到你本地仓库.这样在你下次使用的时候就不需要从远程下载了.如果你所需 ...

  2. win8以管理员身份安装软件

    win8以管理员身份安装软件 msiexec /package

  3. HTML5 中已经可以用 Ajax 上传文件了,而且代码非常简单,借助 FormData 类即可发送文件数据。

    <?phpif (isset($_POST['upload'])) { var_dump($_FILES); move_uploaded_file($_FILES['upfile']['tmp_ ...

  4. 【hibernate】之标注枚举类型@Enumerated(转载)

    实体Entity中通过@Enumerated标注枚举类型,例如将CustomerEO实体中增加一个CustomerType类型的枚举型属性,标注实体后的代码如下所示. @Entity @Table(n ...

  5. Windows 7 的系统文件修复:sfc /scannow

    在线检查与修复 C:\Windows\system32>sfc /scannow 开始系统扫描.此过程将需要一些时间. 开始系统扫描的验证阶段. 验证 100% 已完成. Windows 资源保 ...

  6. class 文件与dex文件区别 (dvm与jvm区别)及Android DVM介绍

    区别一:dvm执行的是.dex格式文件  jvm执行的是.class文件   android程序编译完之后生产.class文件,然后,dex工具会把.class文件处理成.dex文件,然后把资源文件和 ...

  7. JQuery 常用方法基础教程

    本文转自(http://www.cnblogs.com/Leo_wl/archive/2010/06/22/1762401.html) 对于学习使用jquery 的朋友,能用的到,简单的了解下jque ...

  8. Servlet概述及其生命周期

    Servlet和传统CGI程序相比的优点:   1. 只需要启动一个操作系统进程以及加载一个JVM,大大降低了系统的开销 2. 如果多个请求需要做同样处理的时候,这时只需要加载一个类,这也大大降低了开 ...

  9. storm的作业单元:Topology

    Storm系统的数据处理应用单元,是被打包的被称为Topology的作业. 它是由多个数据处理阶段组合而成的,而每个处理阶段在构造时被称为组件(Component),在运行时被称为任务. 那么,组件根 ...

  10. PLSQL_PLSQL Hint用法总结(概念)

    2014-06-20 Created By BaoXinjian