【白书之路】 232 - Crossword Answers 字符串单词提取

232 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 (1 ≤ r ≤ 10 and
1 ≤ c ≤ 10), 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

又是一道英语阅读题，明明题意很简单，却非要描述的这么麻烦。。。

题意：首先对矩阵中的白格进行编号，编号的顺序是从左到右，从上到下，并且只对合格的白格进行编号。

合格的白格的定义是 水平或者垂直方向的起始格，对于水平方向，起始格是左边为空或者左边为黑格的格子；对于垂直方向，起始格是上面为空或者上面为黑格的格子。

编号完成后，我们需要将其中的单词提取出来，首先提取行中的单词，这些单词以*为分隔符，并且记录这个单词首字母的编号；然后提取列中的单词，这些单词同样是以*为分隔符，并且记录这个单词首字母的编号。

最后输出的时候，需要按照单词的编号从小到大输出，还有最最重要的，两组测试数据之间有一个空行。

这里有点麻烦的是如何将这些单词提取出来，我的方法是扫描一行（列），并且设置一个flag，当遇到一个字母，如果flag=0，说明这个字母是首字母，记录位置，设置flag=1，将该字母送入缓冲区；当遇到*或者行（列）尾的时候，如果flag=1，表示这是一个单词的结尾，从缓冲区保存单词，清空缓冲区，设置flag=0。最后进行排序即可。

#include <iostream>#include <stdio.h>#include <ctype.h>#include <string.h>#include <algorithm>#define MAX 12using namespace std;struct Word{    int corr;//编号    char word[MAX];//单词};bool cmp(const Word &a,const Word b){    return a.corr<b.corr;}char grid_c[MAX][MAX];//保存字母表int grid_i[MAX][MAX];//保存编号表int r,c;Word word[100*MAX];//保存识别出的字母char temp[MAX];//缓冲区int length;//缓冲区长度int num;//识别出的字母个数bool eligible(int i,int j)//判断是否是合格的格子{    if((j-1==-1)||grid_c[i][j-1]=='*')        return true;    else if((i-1==-1)||grid_c[i-1][j]=='*')        return -1;    else        return false;}void init_grid()//对矩阵进行编号{    int i,j,num=1;    for(i=0;i<r;i++)    {        for(j=0;j<c;j++)        {            if(isalpha(grid_c[i][j])&&eligible(i,j))            {                grid_i[i][j]=num++;            }        }    }}void show()//显示{    for(int i=0;i<num;i++)    {        printf("%3d.%s\n",word[i].corr,word[i].word);    }}void get_across()//水平单词{    printf("Across\n");    int flag=0;    num=0;    length=0;    for(int i=0;i<r;i++)    {        for(int j=0;j<c;j++)        {            if(isalpha(grid_c[i][j]))//遇到字母            {                if(flag==0)//字母头部                    word[num].corr=grid_i[i][j];                flag=1;                temp[length++]=grid_c[i][j];            }            if((grid_c[i][j]=='*'||(j==c-1))&&flag==1)//字母结尾            {                temp[length]='\0';                strcpy(word[num++].word,temp);                flag=0;                length=0;            }        }    }    sort(word,word+num,cmp);    show();}void get_down()//垂直字母{    printf("Down\n");    int flag=0;    num=0;    length=0;    for(int j=0;j<c;j++)    {        for(int i=0;i<r;i++)        {            if(isalpha(grid_c[i][j]))            {                if(flag==0)                   word[num].corr=grid_i[i][j];                flag=1;                temp[length++]=grid_c[i][j];            }            if((grid_c[i][j]=='*'||(i==r-1))&&flag==1)            {                temp[length]='\0';                strcpy(word[num++].word,temp);                flag=0;                length=0;            }        }    }    sort(word,word+num,cmp);    show();}int main(){    int i;    int cas=1;    while(scanf("%d",&r)&&r)    {        scanf("%d",&c);        for(i=0;i<r;i++)        {            scanf("%s",grid_c[i]);        }        init_grid();        //printf("hehe");        if(cas!=1)            printf("\n");        printf("puzzle #%d:\n",cas++);        get_across();        get_down();    }    return 0;}