uva 10010 - Where's Waldorf?

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

代码：

#include<stdio.h>

#include<string.h>

int n,m;

char s[51][51],word[20][51];

int dx[8]={-1,-1,0,1,1,1,0,-1};

int dy[8]={0,1,1,1,0,-1,-1,-1};

int isword(int i,int j,int z,char word[51])

{

 int len,t;

 len=strlen(word);

 for(t=1;t<len;t++)

 {

  if(dx[z]*t+i <0 || dx[z]*t+i >=n || dy[z]*t+j <0 || dy[z]*t+j >=m)return 0;

  if(s[dx[z]*t+i][dy[z]*t+j]!=word[t])return 0;

 }

 return 1;

}

int main()

 {

    int i,j,z,k,x,y,num,cases,len,flag;

 scanf("%d",&cases);

    while(cases--)

    {

  scanf("%d %d",&n,&m);

  for(i=0;i<n;i++)

   scanf("%s",s[i]);

  scanf("%d",&num);

  for(i=0;i<num;i++)

   scanf("%s",word[i]);

  for(i=0;i<n;i++)

   for(j=0;j<m;j++)

    if(s[i][j]>='A'&&s[i][j]<='Z')

     s[i][j]=s[i][j]-'A'+'a';

  for(i=0;i<num;i++)

  {

   len=strlen(word[i]);

   for(j=0;j<len;j++)

    if(word[i][j]>='A'&&word[i][j]<='Z')

     word[i][j]=word[i][j]-'A'+'a';

  }

  

  for(k=0;k<num;k++)

  {

   flag=0;

   for(i=0;i<n;i++)

   {

    for(j=0;j<m;j++)

    {

     if(s[i][j]==word[k][0])

     {

      for(z=0;z<8;z++)

      {

       if(isword(i,j,z,word[k])==1)

       {

        flag=1;

        x=i+1;

        y=j+1;

        break;

       }

      }

     }

     if(flag==1)break;

    }

    if(flag==1)break;

   }

   printf("%d %d\n",x,y);

  }

  if(cases)printf("\n");

 }

  return 0;

}