UVa 232 Crossword Answers
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 ccharacters 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
这个问题,注意前面的号码就好了。
附上AC代码:
- #include<stdio.h>
- #include<string.h>
- #include<ctype.h>
- struct As
- {
- int x,y;
- int Num;
- }A[];
- struct Dn
- {
- int x,y;
- int Num;
- }D[];
- int a,b;
- char str[][];
- int t1[][];
- int t2[][];
- int k;
- int i1,i2;
- int Flag;
- int main()
- {
- Flag=;
- while(scanf("%d",&a)!=EOF&&a)
- {
- scanf("%d",&b);
- for(int i=;i<;i++)
- for(int j=;j<;j++)
- {
- t1[i][j]=;
- t2[i][j]=;
- }
- for(int i=;i<a;i++)
- scanf("%s",str[i]);
- for(int i=;i<a;i++)
- for(int j=;j<b;j++)
- {
- if(isalpha(str[i][j]))
- {
- if(i==||j==)
- t1[i][j]=;
- else if(str[i-][j]=='*'||str[i][j-]=='*')
- t1[i][j]=;
- }
- }
- k=;
- i1=i2=;
- for(int i=;i<a;i++)
- for(int j=;j<b;j++)
- {
- if(t1[i][j])
- {
- k++;
- t2[i][j]=k;
- if(i==||str[i-][j]=='*')
- {
- D[i1].Num=k;
- D[i1].x=i;
- D[i1++].y=j;
- }
- if(j==||str[i][j-]=='*')
- {
- A[i2].Num=k;
- A[i2].x=i;
- A[i2++].y=j;
- }
- }
- }
- if(Flag!=) printf("\n");
- printf("puzzle #%d:\n",Flag++);
- printf("Across\n");
- for(int i=;i<i2;i++)
- {
- printf("%3d.",A[i].Num);
- for(int j=;;j++)
- {
- printf("%c",str[A[i].x][A[i].y+j]);
- if((A[i].y+j)==b-) break;
- if(str[A[i].x][A[i].y+j+]=='*') break;
- }
- printf("\n");
- }
- printf("Down\n");
- for(int i=;i<i1;i++)
- {
- printf("%3d.",D[i].Num);
- for(int j=;;j++)
- {
- printf("%c",str[D[i].x+j][D[i].y]);
- if((D[i].x+j)==a-) break;
- if(str[D[i].x+j+][D[i].y]=='*') break;
- }
- printf("\n");
- }
- }
- return ;
- }
修改之后的代码:
速度快了一点,代码长了一点,谢谢 garbageMan
- #include<stdio.h>
- #include<string.h>
- #include<ctype.h>
- struct As
- {
- int x,y;
- int Num;
- } A[];
- struct Dn
- {
- int x,y;
- int Num;
- } D[];
- int Sum1,Sum2;
- char str[][];
- int t1[][];
- int t2[][];
- void Initializer();
- void Input(int a);
- void SearchHead(int ,int );
- void OrderHead(int ,int );
- void SortAorD(int ,int );
- void Output(int ,int ,int );
- int main()
- {
- int Flag=;
- int a,b;
- while(scanf("%d",&a)!=EOF&&a)
- {
- Sum1=Sum2=;
- scanf("%d",&b);
- Initializer();
- Input(a);
- SearchHead(a,b);
- OrderHead(a,b);
- SortAorD(a,b);
- Output(Flag,a,b);
- Flag++;
- }
- return ;
- }
- void Initializer()
- {
- for(int i=; i<; i++)
- for(int j=; j<; j++)
- {
- t1[i][j]=;
- t2[i][j]=;
- }
- }
- void Input(int a)
- {
- for(int i=; i<a; i++)
- scanf("%s",str[i]);
- }
- void SearchHead(int a,int b)
- {
- for(int i=; i<a; i++)
- for(int j=; j<b; j++)
- {
- if(isalpha(str[i][j]))
- {
- if(i==||j==)
- t1[i][j]=;
- else if(str[i-][j]=='*'||str[i][j-]=='*')
- t1[i][j]=;
- }
- }
- }
- void OrderHead(int a,int b)
- {
- for(int i=; i<a; i++)
- for(int j=; j<b; j++)
- {
- if(isalpha(str[i][j]))
- {
- if(i==||j==)
- t1[i][j]=;
- else if(str[i-][j]=='*'||str[i][j-]=='*')
- t1[i][j]=;
- }
- }
- }
- void SortAorD(int a,int b)
- {
- int k=;
- for(int i=; i<a; i++)
- for(int j=; j<b; j++)
- {
- if(t1[i][j])
- {
- k++;
- t2[i][j]=k;
- if(i==||str[i-][j]=='*')
- {
- D[Sum1].Num=k;
- D[Sum1].x=i;
- D[Sum1++].y=j;
- }
- if(j==||str[i][j-]=='*')
- {
- A[Sum2].Num=k;
- A[Sum2].x=i;
- A[Sum2++].y=j;
- }
- }
- }
- }
- void Output(int Flag,int a,int b)
- {
- if(Flag!=) printf("\n");
- printf("puzzle #%d:\n",Flag);
- printf("Across\n");
- for(int i=; i<Sum2; i++)
- {
- printf("%3d.",A[i].Num);
- for(int j=;; j++)
- {
- printf("%c",str[A[i].x][A[i].y+j]);
- if((A[i].y+j)==b-) break;
- if(str[A[i].x][A[i].y+j+]=='*') break;
- }
- printf("\n");
- }
- printf("Down\n");
- for(int i=; i<Sum1; i++)
- {
- printf("%3d.",D[i].Num);
- for(int j=;; j++)
- {
- printf("%c",str[D[i].x+j][D[i].y]);
- if((D[i].x+j)==a-) break;
- if(str[D[i].x+j+][D[i].y]=='*') break;
- }
- printf("\n");
- }
- }
UVa 232 Crossword Answers的更多相关文章
- UVa232.Crossword Answers
题目链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...
- 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:htt ...
- poj 1888 Crossword Answers 模拟题
Crossword Answers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 869 Accepted: 405 D ...
- Crossword Answers UVA - 232
题目大意 感觉挺水的一道题.找出左面右面不存在或者是黑色的格子的白各,然后编号输出一横向单词和竖向单词(具体看原题) 解析 ①找出各个格子的编号 ②对每个节点搜索一下 ③输出的时候注意最后一个数据后面 ...
- 【习题 3-6 UVA - 232】Crossword Answers
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 模拟题.注意场宽为3 [代码] #include <bits/stdc++.h> using namespace std ...
- 紫书第三章训练1 D - Crossword Answers
A crossword puzzle consists of a rectangular grid of black and white squares and two lists of defini ...
- 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 ...
- Crossword Answers -------行与列按序输出
题目链接:https://vjudge.net/problem/UVA-232#author=0 题意:关键句:The de nitions correspond to the rectangular ...
- [刷题]算法竞赛入门经典 3-4/UVa455 3-5/UVa227 3-6/UVa232
书上具体所有题目:http://pan.baidu.com/s/1hssH0KO 题目:算法竞赛入门经典 3-4/UVa455:Periodic Strings 代码: //UVa455 #inclu ...
随机推荐
- caffe绘制训练过程的loss和accuracy曲线
转自:http://blog.csdn.net/u013078356/article/details/51154847 在caffe的训练过程中,大家难免想图形化自己的训练数据,以便更好的展示结果.如 ...
- 用tee和script来记录终端输出
如何在输出信息的同时把数据保存到文件当中? 一,如何把命令运行的结果保存到文件当中?这个问题太简单了,大家都知道,用 > 把输出转向就可以了 例子:[lhd@hongdi ~]$ ls > ...
- [LeetCode 121] - 买入与卖出股票的最佳时机(Best Time to Buy and Sell Stock)
问题 假设你有一个数组,其中的第i个元素表示一只股票在第i天的价格. 如果只允许你完成一次交易(即买入并卖出股票一次),设计一个找出最大利润的算法. 初始思路 和122一样,基于买入与卖出股票的最佳时 ...
- CCI_chapter 8 Recurision
8.1 水题 8.2 Imagine a robot sitting on the upper left hand corner of an NxN grid The robot can only m ...
- 二极管IN4001~IN4007参数
电压范围50~1000V 正向导通电流1A 导通电压降:1.1V 具体见下图:
- Qt编程之通过鼠标滚轮事件缩放QGraphicsView里面的Item
首先自己subclass QGraphicsView的一个类,叫DiagramView,然后重新实现它的滚轮事件函数,然后发送一个缩放信号: oid DiagramView::wheelEvent(Q ...
- VS环境下的makefile编译
直接找这个了,原来VS也可以makefile,在windows上解析makefile的软件叫NMAKE.exe 打算用命令Cmake -G“NMake Makefiles” 生成VS环境下Nmake的 ...
- RedisTemplate
Spring Boot中Jedis几个api返回值的确认 @RequestMapping("/del/{key}") public String del(@PathVariable ...
- Java代码编译和执行的整个过程
Java代码编译是由Java源码编译器来完成,流程图如下所示: Java字节码的执行是由JVM执行引擎来完成,流程图如下所示: Java代码编译和执行的整个过程包含了以下三个重要的机制: Java源码 ...
- ajax提交写法
<script> /* ajax提交写法 */ function add_prize() { // var query={}; var query = new Object(); quer ...