Uva 232 一个换行WA 了四次
由于UVA OJ上没有Wrong anwser,搞的多花了好长时间去测试程序,之前一直以为改OJ有WA,后来网上一搜才知道没有WA,哎哎浪费了好长时间。此博客用来记录自己的粗心大意。
链接地址:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=4&page=show_problem&problem=168
232 - Crossword AnswersTime limit: 3.000 seconds |
|
Crossword Answers A crossword puzzle consists of a rectangular grid of black and white squares and two lists of definitions (or descriptions). One list of definitions is for ``words" to be written left to right across white squares in the rows and the other list is for words to be written down white squares in the columns. (A word is a sequence of alphabetic characters.) To solve a crossword puzzle, one writes the words corresponding to the definitions on the white squares of the grid. The definitions correspond to the rectangular grid by means of sequential integers on ``eligible" white squares. White squares with black squares immediately to the left or above them are ``eligible." White squares with no squares either immediately to the left or above are also ``eligible." No other squares are numbered. All of the squares on the first row are numbered. The numbering starts with 1 and continues consecutively across white squares of the first row, then across the eligible white squares of the second row, then across the eligible white squares of the third row and so on across all of the rest of the rows of the puzzle. The picture below illustrates a rectangular crossword puzzle grid with appropriate numbering. An ``across" word for a definition is written on a sequence of white squares in a row starting on a numbered square that does not follow another white square in the same row. The sequence of white squares for that word goes across the row of the numbered square, ending immediately before the next black square in the row or in the rightmost square of the row. A ``down" word for a definition is written on a sequence of white squares in a column starting on a numbered square that does not follow another white square in the same column. The sequence of white squares for that word goes down the column of the numbered square, ending immediately before the next black square in the column or in the bottom square of the column. Every white square in a correctly solved puzzle contains a letter. You must write a program that takes several solved crossword puzzles as input and outputs the lists of across and down words which constitute the solutions. Input Each puzzle solution in the input starts with a line containing two integers r and c ( and ), where r (the first number) is the number of rows in the puzzle and c (the second number) is the number of columns. The r rows of input which follow each contain c characters (excluding the end-of-line) which describe the solution. Each of those c characters is an alphabetic character which is part of a word or the character ``*", which indicates a black square. The end of input is indicated by a line consisting of the single number 0. Output Output for each puzzle consists of an identifier for the puzzle (puzzle #1:, puzzle #2:, etc.) and the list of across words followed by the list of down words. Words in each list must be output one-per-line in increasing order of the number of their corresponding definitions. The heading for the list of across words is ``Across". The heading for the list of down words is ``Down". In the case where the lists are empty (all squares in the grid are black), the Across and Down headings should still appear. Separate output for successive input puzzles by a blank line. Sample Input 2 2 AT *O 6 7 AIM*DEN *ME*ONE UPON*TO SO*ERIN *SA*OR* IES*DEA 0 Sample Output puzzle #1: Across 1.AT 3.O Down 1.A 2.TO puzzle #2: Across 1.AIM 4.DEN 7.ME 8.ONE 9.UPON 11.TO 12.SO 13.ERIN 15.SA 17.OR 18.IES 19.DEA Down 1.A 2.IMPOSE 3.MEO 4.DO 5.ENTIRE 6.NEON 9.US 10.NE 14.ROD 16.AS 18.I 20.A |
源代码如下:
#include<iostream>
#include<string.h>
using namespace std;
#define SIZE 20
#define MAX 200
int main()
{ int pu[SIZE][SIZE];
char s[SIZE][SIZE];
char str[MAX][SIZE];
int r, c;
int T=0; while(cin>>r && r!= 0 )
{
int count=1;
for(int i=0; i<MAX; i++)
{
for(int j=0; j<SIZE ; j++)
str[i][j]='\0';
}
cin>>c;
for(int i=0; i<r ; i++)
{
for(int j=0; j<c ; j++)
{ pu[i][j]=0;
cin>>s[i][j]; if( i==0 ) //diyihang
{
if(s[i][j] != '*')
{
pu[i][j]=count;
count++;
}
// else s[i][j]=0; }
else
{
if( j==0 && s[i][j] != '*')
{ pu[i][j]=count;
count++;
}
else if(s[i][j-1]=='*' && s[i][j] != '*')
{
pu[i][j]= count;
count++;
}
else if(s[i-1][j]=='*' && s[i][j] != '*')
{
pu[i][j]= count;
count++;
}
else ;
} }
} if(T >0 ) cout<<endl; //我之前在后面直接cout<<endl,这样便会造成会多输出一空行,造成WA,所以一定要切记!
T++;
cout<<"puzzle #"<<T<<":"<<endl;
cout<<"Across"<<endl;
for(int i=0; i<r ; i++)
{ bool flag = false;
int start=1;
// bool sflag= false;
for(int j=0; j<c ; j++)
{
if( s[i][j] == '*')
{ if(flag)
{
cout<<endl;
flag = false;
}
}
else
{ if( !flag )
{
start =pu[i][j];
if(start <10 )
cout<<" ";
else cout<<" ";
cout<<start<<".";
flag = true;
}
cout<<s[i][j];
// flag = true;
if( j== c-1)
{
cout<<endl;
flag = false;
}
} } }
// cout<<"Down"<<endl;
int locstr=0;
// int start=1;
for(int j=0; j<c ; j++)
{ bool flag = false; int countstr=0;
for(int i=0; i<r ; i++)
{ if( s[i][j] =='*' )
{
if(flag)
{
// cout<<endl;
// flag = false;
str[locstr][countstr]='\0';
// locstr++;
countstr=0;
flag = false;
} }
else
{
if( !flag )
{ locstr =pu[i][j]; flag = true;
// str[locstr][countstr++]=""
/* start =pu[i][j];
if(start <10 )
cout<<" ";
else cout<<" ";
cout<<start<<".";
flag = true;
*/
}
// cout<<s[i][j];
str[ locstr][countstr++]=s[i][j];
// flag = true;
if( i== r-1)
{
// cout<<endl;
// flag = false;
str[locstr][countstr]='\0';
// locstr++;
countstr=0;
flag = false;
} } } }
// qsort(str,MAX,sizeof(str[0]),comp);
cout<<"Down"<<endl;
for(int i=0; i<MAX; i++)
{
int len = strlen(str[i]);
if(len >0 )
{ if(i <10 )
cout<<" ";
else cout<<" ";
cout<<i<<".";
cout<<str[i]<<endl;
// cout<<start<<"."; } } // cout<<endl;
} return 0;
}
Uva 232 一个换行WA 了四次的更多相关文章
- Qt applendPlainText()/append() 多添加一个换行解决方法
Qt applendPlainText()/append() 多添加一个换行解决方法 void ConsoleDialog::appendMessageToEditor(const QString & ...
- print默认在末尾添加一个换行符,但其实也可以不用!
白月黑羽今天给大家分享另外一个冷知识:) python中 print函数打印信息时默认添加一个换行符,所以你看到一条print语句,则单独占一行,那是否可以不要换行符呢? 答案是肯定的,当然可以,可用 ...
- express + mongodb 搭建一个简易网站 (四)
express + mongodb 搭建一个简易网站 (四) 目前网站整体页面都已经能全部展示了,但是,整个网站还有两个块需要做完才能算完整,一个连接数据库,目前网站上的数据都是抓取的本地假数据,所以 ...
- JavaScript确定一个字符串是否包含在另一个字符串中的四种方法
一.indexOf() 1.定义 indexOf()方法返回String对象第一次出现指定字符串的索引,若未找到指定值,返回-1.(数组同一个概念) 2.语法 str.indexOf(searchVa ...
- 利用django创建一个投票网站(四)
创建你的第一个 Django 项目, 第四部分 这一篇从第三部分(zh)结尾的地方继续讲起.我们将继续编写投票应用,专注于简单的表单处理并且精简我们的代码. 编写一个简单的表单 让我们更新一下在上一个 ...
- 用C写一个web服务器(四) CGI协议
* { margin: 0; padding: 0 } body { font: 13.34px helvetica, arial, freesans, clean, sans-serif; colo ...
- UVA 232 Corssword Answer
题意:输入m*n大小的字符串(里面有*,*为黑格,其他为白格),然后对它编号,编号规则为从左到右,从上往下,且左边或上面没有白格(可能是黑格或越界),如下图: 注意: ①除第一次输出答案外,其余每次输 ...
- Play Framework 完整实现一个APP(十四)
添加测试 ApplicationTest.java @Test public void testAdminSecurity() { Response response = GET("/adm ...
- 调度系统任务创建---创建一个MultiJob的任务(四)
我们如果下面这种拓扑结构的调度任务,该任务的拓扑结构就是一个有向五环图DAG,有fork,有join的操作等. 可以通过jenkins创建MultiJob的任务实现: 实例任务的拓扑结构: Multi ...
随机推荐
- C#读取Txt大数据并更新到数据库
环境 Sqlserver 2016 .net 4.5.2 目前测试数据1300万 大约3-4分钟.(限制一次读取条数 和 线程数是 要节省服务器资源,如果调太大服务器其它应用可能就跑不了了), Sql ...
- CRNN+CTC (基于CTPN 的end-to-end OCR)
1. https://zhuanlan.zhihu.com/p/43534801 (详细原理) 2. https://blog.csdn.net/forest_world/article/detai ...
- 《Head First 设计模式》笔记
第一章 策略模式 00设计原则:找出应用中可能需要变化之处,把它们独立出来,不要和那些不需要变化的代码放在一起. 把会变化的部分取出并封装起来,好让其它部分不会受到影响.结果如何?代码变化引起的不经意 ...
- Java并发编程知识点总结Volatile、Synchronized、Lock实现原理
Volatile关键字及其实现原理 在多线程并发编程中,Volatile可以理解为轻量级的Synchronized,用volatile关键字声明的变量,叫做共享变量,其保证了变量的“可见性”以及“有序 ...
- unity shader之预备知识
1.渲染流水线 任务:从一个三维场景出发,生成(或者渲染)一张二维图像.即:计算机需要从一系列的定点出数据,纹理等信息出发,把这些信息最终转换程一张人眼可以看到的图像.而这个工作通常是由CPU和GPU ...
- css3的@media
都知道bootstrap响应式布局很酷,但是是怎么实现的呢?其官网首页有提到这一切的功劳都是来自于CSS 媒体查询(Media Query). 使用 @media 查询,你可以针对不同的媒体类型定义不 ...
- iframe中使用模态框提交表单后,iframe加载父页面的解决方法
在iframe中使用模态框提交表单后,会出现iframe加载整个父页面的问题,如下图: 解决方法: 在form表单中添加target属性 _parent 这个属性会使目标文档载入父窗口或者包含来超链接 ...
- 【管理学】SMART
- CF 551 D.Serval and Rooted Tree 树形DP
传送门:http://codeforces.com/contest/1153/problem/D 思路: 这道题想了一天,突发奇想,就是维护每个点两个值,第几大和第几小,就可以有传递性了. #incl ...
- HDU - 2255 奔小康赚大钱 KM算法 模板题
HDU - 2255 题意: 分配n所房子给n个家庭,不同家庭对一所房子所需缴纳的钱是不一样的,问你应当怎么分配房子,使得最后收到的钱最多. 思路: KM算法裸题.上模板 #include <i ...