UVA 10010

Given a m by n grid of letters, ( ), and a list of words, find thelocation in the grid at which the word can be found. A word matchesa straight, uninterrupted line of letters in the grid. A word canmatch the letters in the grid regardless of case (i.e. upper
andlower case letters are to be treated as the same). The matching canbe done in any of the eight directions either horizontally,vertically or diagonally through the grid.

Input

The input begins with a single positive integer on a line by itselfindicating the number of the cases following, each of them asdescribed below. This line is followed by a blank line, and thereis also a blank line between two consecutive inputs.

The input begins with a pair of integers, m followed byn，1<=m,n<=50, in decimal notation ona single line. The next m lines contain n letters each; this is thegrid of letters in which the words of the list must be found. Theletters in the grid may be in upper
or lower case. Following thegrid of letters, another integer k appears on a line by itself (1<=k<=20). The next k lines of inputcontain the list of words to search for, one word per line. Thesewords may contain upper and lower case letters only (no spaces,hyphens
or other non-alphabeticcharacters). 

Output

For each test case, the output must follow the description below.The outputs of two consecutive cases will be separated by a blankline.

For each word in the word list, a pair of integers representing thelocation of the corresponding word in the grid must be output. Theintegers must be separated by a single space. The first integer isthe line in the grid where the first letter of the given word
canbe found (1 represents the topmost line in the grid, and mrepresents the bottommost line). The second integer is the columnin the grid where the first letter of the given word can be found(1 represents the leftmost column in the grid, and n represents therightmost
column in the grid). If a word can be found more thanonce in the grid, then the location which is output shouldcorrespond to the uppermost occurence of the word (i.e. theoccurence which places the first letter of the word closest to thetop of the grid). If
two or more words are uppermost, the outputshould correspond to the leftmost of these occurences. All wordscan be found at least once in the grid.

Sample Input

1

8 11

abcDEFGhigg

hEbkWalDork

FtyAwaldORm

FtsimrLqsrc

byoArBeDeyv

Klcbqwikomk

strEBGadhrb

yUiqlxcnBjf

4

Waldorf

Bambi

Betty

Dagbert

Sample Output

2 5

2 3

1 2

7 8

代码

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define N 60
int search(char list[]
, char word[], int row, int col, int m, int n, int len)
{
int i, j, k;
for (i = row, k = 0; i >= 0 && k < len && list[i][col] == word[k]; --i, ++k) ;
if (k == len) return 1;
for (i = row, k = 0; i < m && k < len && list[i][col] == word[k]; ++i, ++k) ;
if (k == len) return 1;
for (j = col, k = 0; j >= 0 && k < len && list[row][j] == word[k]; --j, ++k) ;
if (k == len) return 1;
for (j = col, k = 0; j < n && k < len && list[row][j] == word[k]; ++j, ++k) ;
if (k == len) return 1;
for (i = row, j = col, k = 0; i >= 0 && j >= 0 && k < len && list[i][j]
== word[k]; i--, j--, ++k) ;
if (k == len) return 1;
if (k == len) return 1;
for (i = row, j = col, k = 0; i >= 0 && j < n && k < len && list[i][j] == word[k]; i--, j++, ++k) ;
if (k == len) return 1;
for (i = row, j = col, k = 0; i < m && j >= 0 && k < len && list[i][j] == word[k]; i++, j--, ++k) ;
if (k == len) return 1;
for (i = row, j = col, k = 0; i < m && j < n && k < len && list[i][j] == word[k]; i++, j++, ++k) ;
return k == len;
}
int main()
{
char list

, words

;
int i, j, k, z, m, n, cases;;
scanf("%d", &cases);
while (cases--) {
int found
= {0};
scanf("%d%d", &m, &n);
for (i = 0; i < m; ++i)
scanf("%s", list[i]);
for (i = 0; i < m; ++i)
for (j = 0; j < n; ++j)
list[i][j] = toupper(list[i][j]);
scanf("%d", &k);
for (i = 0; i < k; ++i)
scanf("%s", words[i]);
for (i = 0; i < k; ++i)
for (j = 0; j < strlen(words[i]); ++j)
words[i][j] = toupper(words[i][j]);
for (z = 0; z < k; ++z)
for (i = 0; i < m; ++i)
for (j = 0; j < n; ++j)
if (!found[z] && search(list, words[z], i, j, m, n, strlen(words[z]))) {
printf("%d %d\n", i+1, j+1);
found[z] = 1;
}
if (cases)
printf("\n");
}
return 0;
}