232:Crossword Answers

Crossword Answers

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 1357		Accepted: 572

Description

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.

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

Source

World Finals 1994

自己的代码测试数据没有问题，各种输出格式也都检查了，一直WA。。。最后才发现题意理解有偏差，但是题目给的测试用例正好能通过。。。其实Across的话就是从按行从上到下输出就行了，列也是一样，我还以为是先输出首字母在第一行的然后是第二行...这样，找了好久的Bug，还是要好好审题啊。。。

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 10 + 5;
char table[maxn][maxn];
int cnts[maxn][maxn];
int r,c;
bool isRe(int x,int y){
    if(!y || table[x][y-1] == '*') return 1;
    return 0;
}
bool isCe(int x,int y){
    if(!x || table[x-1][y] == '*') return 1;
    return 0;
}
void Aoutput(int x,int y){
    printf("%3d.",cnts[x][y]);
    for(int i = y;i < c && table[x][i] != '*';i++){
        printf("%c",table[x][i]);
    }
    putchar('\n');
}
void Doutput(int x,int y){
    printf("%3d.",cnts[x][y]);
    for(int i = x;i < r && table[i][y] != '*';i++){
        printf("%c",table[i][y]);
    }
    putchar('\n');
}
int main(){
    int puzzle = 0;
    while(scanf("%d",&r) && r){
        scanf("%d",&c);
        memset(table,'\0',sizeof(table));
        memset(cnts,0,sizeof(cnts));
        for(int i = 0;i < r;i++){
            scanf("%s",table[i]);
        }
        int cnt = 0;
        for(int i = 0;i < r;i++){
            for(int j = 0;j < c;j++){
                if(table[i][j] != '*' && (isRe(i,j) || isCe(i,j))){
                    cnts[i][j] = ++cnt;
                }
            }
        }
        if(++puzzle > 1) putchar('\n');
        printf("puzzle #%d:\n",puzzle);
        printf("Across\n");
        cnt = 0;
        for(int i = 0;i < r;i++){
            for(int j = 0;j < c;j++){
                if(cnts[i][j] && isRe(i,j)) Aoutput(i,j);
            }
        }
        printf("Down\n");
        cnt = 0;
        for(int i = 0;i < r;i++){
            for(int j = 0;j < c;j++){
                if(cnts[i][j] && isCe(i,j)) Doutput(i,j);
            }
        }
    }
    return 0;
}

上面的代码有很多重复判断，所以效率不高，改了一下：

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 10 + 5;
char table[maxn][maxn];
int cnts[maxn][maxn];
int row[maxn][maxn];
int com[maxn][maxn];
int r,c;
bool isRe(int x,int y){
    if(!y || table[x][y-1] == '*') return 1;
    return 0;
}
bool isCe(int x,int y){
    if(!x || table[x-1][y] == '*') return 1;
    return 0;
}
void Aoutput(int x,int y){
    printf("%3d.",cnts[x][y]);
    for(int i = y;i < c && table[x][i] != '*';i++){
        printf("%c",table[x][i]);
    }
    putchar('\n');
}
void Doutput(int x,int y){
    printf("%3d.",cnts[x][y]);
    for(int i = x;i < r && table[i][y] != '*';i++){
        printf("%c",table[i][y]);
    }
    putchar('\n');
}
int main(){
    int puzzle = 0;
    while(scanf("%d",&r) && r){
        scanf("%d",&c);
        memset(table,'\0',sizeof(table));
        memset(cnts,0,sizeof(cnts));
        memset(row,0,sizeof(row));
        memset(com,0,sizeof(com));
        for(int i = 0;i < r;i++){
            scanf("%s",table[i]);
        }
        int cnt = 0;
        for(int i = 0;i < r;i++){
            for(int j = 0;j < c;j++){
                if(table[i][j] != '*'){
                    row[i][j] = isRe(i,j);
                    com[i][j] = isCe(i,j);
                    if(row[i][j] + com[i][j]) cnts[i][j] = ++cnt;
                }
            }
        }
        if(++puzzle > 1) putchar('\n');
        printf("puzzle #%d:\n",puzzle);
        printf("Across\n");
        cnt = 0;
        for(int i = 0;i < r;i++){
            for(int j = 0;j < c;j++){
                if(row[i][j]) Aoutput(i,j);
            }
        }
        printf("Down\n");
        cnt = 0;
        for(int i = 0;i < r;i++){
            for(int j = 0;j < c;j++){
                if(com[i][j]) Doutput(i,j);
            }
        }
    }
    return 0;
}

时间减少了差不多20ms，还是不高。。。

这位博主是扩展网格边界，可以简化程序逻辑，使得边缘格的处理逻辑与中间格的处理逻辑相同。使得判断网格是否为单词的开始网格条件统一；同时，计算行单词和列单词的逻辑也变得简单。他的代码是32ms。

/* UVA232 UVALive5171 POJ1888 Crossword Answers */

#include <stdio.h>

#define MAXN 10

char grid[MAXN+2][MAXN+2];

/* 输出行单词 */
void rowoutput(int no, int row, int col)
{
    printf("%3d.", no);
    while(grid[row][col] != '*')
        putchar(grid[row][col++]);
    putchar('\n');
}

/* 输出列单词 */
void coloutput(int no, int row, int col)
{
    printf("%3d.", no);
    while(grid[row][col] != '*')
        putchar(grid[row++][col]);
    putchar('\n');
}

int main(void)
{
    int r, c, caseno=0, wcount, i, j;

    while(scanf("%d", &r) != EOF && r != 0) {
        /* 已经读入行，继续读入列 */
        scanf("%d", &c);
        getchar();

        /* 读入网格 */
        for(i=1; i<=r; i++) {
            for(j=1; j<=c; j++)
                grid[i][j] = getchar();
            getchar();      /* '\n' */
        }

        // 设置边界
        for(i=0; i<r+2; i++) {
            grid[i][0] = '*';
            grid[i][c+1] = '*';
        }
        for(i=0; i<c+2; i++) {
            grid[0][i] = '*';
            grid[r+1][i] = '*';
        }

        /* 计算并且输出结果 */
        if(++caseno > 1)
            printf("\n");
        printf("puzzle #%d:\n", caseno);

        /* 行计算并且输出结果 */
        printf("Across\n");
        wcount = 0;
        for(i=1; i<=r; i++) {
            for(j=1; j<=c; j++)
                if(grid[i][j] != '*' && (grid[i-1][j] == '*' || grid[i][j-1] == '*')) {
                    wcount++;
                    if(grid[i][j-1] == '*')
                        rowoutput(wcount, i, j);
                }
        }

        /* 列计算并且输出结果 */
        printf("Down\n");
        wcount = 0;
        for(i=1; i<=r; i++) {
            for(j=1; j<=c; j++)
                if(grid[i][j] != '*' && (grid[i-1][j] == '*' || grid[i][j-1] == '*')) {
                    wcount++;
                    if(grid[i-1][j] == '*')
                        coloutput(wcount, i, j);
                }
        }
    }

    return 0;
}