Where's Waldorf? 
Given a 
m
 by 
n
 grid of letters, ( 
), and a list of words, find the location in the grid at which the word can be found. A word matches a straight, uninterrupted line of letters in the grid. A word can match the letters in the grid regardless of case (i.e. upper and lower case letters are to be treated as the same). The matching can be 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 itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The input begins with a pair of integers, m followed by n,  in decimal notation on a single line. The next m lines contain n letters each; this is the grid of letters in which the words of the list must be found. The letters in the grid may be in upper or lower case. Following the grid of letters, another integer k appears on a line by itself ( ). The next k lines of input contain the list of words to search for, one word per line. These words may contain upper and lower case letters only (no spaces, hyphens or other non-alphabetic characters).

Output 

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
For each word in the word list, a pair of integers representing the location of the corresponding word in the grid must be output. The integers must be separated by a single space. The first integer is the line in the grid where the first letter of the given word can be found (1 represents the topmost line in the grid, and m represents the bottommost line). The second integer is the column in the grid where the first letter of the given word can be found (1 represents the leftmost column in the grid, and n represents the rightmost column in the grid). If a word can be found more than once in the grid, then the location which is output should correspond to the uppermost occurence of the word (i.e. the occurence which places the first letter of the word closest to the top of the grid). If two or more words are uppermost, the output should correspond to the leftmost of these occurences. All words can 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

Miguel Revilla  2000-08-22
이 문제는 WA가 여러 번 나왔는데 처음에는 대소문자를 구분하지 않고 입력한 데이터가 정답을 낼 수 있었다.나중에 알았어요. 수정하고 나서.두 번째는 이 문제의 하표가 1부터 시작된 것을 발견하지 못했다.나중에 수정한 후에도 WA였는데 다른 코드를 보고 두 번째 테스트 때 빈 줄이 있어야 한다는 것을 알게 되었다.앞으로 격식에 주의하세요...이 문제의 코드는 매우 난잡하게 썼다

#include <stdio.h>
#include <string.h>

#define MAXN	100

char grid[MAXN][MAXN];
char word[MAXN][MAXN];

int foo(int x, int y, char word[], int m, int n)
{
	int i, j, k;
	int len;
	int flag;

	len = strlen(word);

	if (len + y < n) {				/* left-right */
		flag = 1;
		for (i=0; i<len; ++i) 
			if (word[i] != grid[x][y+i]) {
				flag = 0;
				break;
			} 
		if (flag)
			return 1;
	}

	if (y - len >= 0) {			/* right-left */
		flag = 1;
		for (i=0; i<len; ++i)
			if (word[i] != grid[x][y-i]) {
				flag = 0;
				break;
			}
		if (flag)
			return 1;
	}

	if (x - len >= 0) {			/* down-up */
		flag = 1;
		for (i=0; i<len; ++i)
			if (word[i] != grid[x-i][y]) {
				flag = 0;
				break;
			}
		if (flag)
			return 1;
	}

	if (x + len < m) {			/* up-down */
		flag = 1;
		for (i=0; i<len; ++i)
			if (word[i] != grid[x+i][y]) {
				flag = 0;
				break;
			}
		if (flag)
			return 1;
	}

	if (y-len+1>=0 && x-len+1>=0) {		/* left-up */
		flag = 1;
		for (i=0; i<len; ++i)
			if (word[i] != grid[x-i][y-i]) {
				flag = 0;
				break;
			}
		if (flag)
			return 1;
	}

	if (y+len-1<n && x-len+1>=0) {		/* right-up */
		flag = 1;
		for (i=0; i<len; ++i)
			if (word[i] != grid[x-i][y+i]) {
				flag = 0;
				break;
			}
		if (flag)
			return 1;
	}

	if (x+len-1<m && y-len+1>=0) {		/* left-down */
		flag = 1;
		for (i=0; i<len; ++i)
			if (word[i] != grid[x+i][y-i]) {
				flag = 0;
				break;
			}
		if (flag)
			return 1;
	}

	if (x+len-1<m && y+len-1<n) {		/* right-down */
		flag = 1;
		for (i=0; i<len; ++i)
			if (word[i] != grid[x+i][y+i]) {
				flag = 0;
				break;
			}
		if (flag)
			return 1;		
	}
	return 0;
}

int main(void)
{
	int count, nword;
	int m, n;
	int i, j, k;
	int flag;
	
	scanf("%d", &count);
	while (count--) {
		scanf("%d%d", &m, &n);	
		for (i=0; i<m; ++i)
			scanf("%s", grid[i]);
		
		scanf("%d", &nword);
		for (i=0; i<nword; ++i)
			scanf("%s", word[i]);
		
		for (i=0; i<m; ++i)
			for (j=0; j<n; ++j)
				if (isupper(grid[i][j]))
					grid[i][j] += 32;
		
		for (i=0; i<nword; ++i)
			for (j=0; j<strlen(word[i]); ++j)
				if (isupper(word[i][j]))
					word[i][j] += 32;

		for (k=0; k<nword; ++k) {
			flag = 0;
			for (i=0; i<m; ++i) {
				for (j=0; j<n; ++j) {
					if (grid[i][j] == word[k][0]) 
						flag = foo(i, j, word[k], m, n);
					if (flag) {
						printf("%d %d
", i+1, j+1);
						goto loop;
					} else
						continue;
				}
			}
			loop:	;
		}
		if (count)
			putchar('
');
	}
}