Where's Waldorf?

Given a m by n grid of letters, (1 ≤ m, n ≤ 50), 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, 1 ≤ m, n ≤ 50 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 (1 ≤ k ≤ 20). 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
题解：遍历每一个点，然后再这个点上下左右等八个方向进行搜索就可以了

#include<stdio.h>#include<string.h>int fun(int ,int ); char a[52][52],s[21],b[20];int next[8][2]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{-1,1},{1,-1},{1,1}};int r,c;int main(){    int n,i,j,k,f;    scanf("%d",&n);    while(n--)    {        scanf("%d%d",&r,&c);        for(i=0;i<r;i++)        {            scanf("%s",a[i]);        }        for(i=0;i<r;i++)        {            for(j=0;j<c;j++)            {                if(a[i][j]>='A'&&a[i][j]<='Z')                a[i][j]=a[i][j]+32;            }        }        scanf("%d",&k);        getchar();        while(k--)        {            scanf("%s",s);            for(j=0;j<strlen(s);j++)            {                if(s[j]>='A'&&s[j]<='Z')                s[j]=s[j]+32;            }                for(i=0;i<r;i++)                {                    f=0;                    for(j=0;j<c;j++)                    {                        if(fun(i,j)==1)                        {                            printf("%d %d\n",i+1,j+1);                            f=1;                            break;                        }                    }                    if(f==1)                    break;                }        }    if(n)     printf("\n");    }    return 0;}int fun(int x,int y){    int tx,ty,i,j,flag,k;    for(i=0;i<8;i++)    {        flag=0; tx=x;ty=y;        for(k=0;k<strlen(s)+1;k++)        b[k]='\0';        for(j=0;j<strlen(s);j++)        {           if(tx<0||tx>=r||ty<0||ty>=c)             break;            b[j]=a[tx][ty];                 tx=tx+next[i][0];            ty=ty+next[i][1];                   }        if(strcmp(b,s)==0)        {            flag=1;            break;        }    }    return flag;}