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 nletters 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

uva的题，坑点比较多 英语也是问题 这题大意是给一个m*n的字符矩阵 然后对于给定的一个单词，在这个字符矩阵里面搜到这个单词最开始出现的位置，（从上往下，从左往右搜）搜索可以有8个方向 如下图

即上下左右以及两条对角线的方向

暴力搜就可以了

#include<cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cctype>
using namespace std;
char s[100][100];
char x[100];
int is_find(int m,int n,int i,int j)//8个方向搜索
{
	int k,p,h,l,flag,len=strlen(x);
	if(i+1>=len)
	{
		p=0;flag=1;
		for(k=i;k>=i+1-len;k--)
			if(s[k][j]!=x[p++])
			{flag=0;break;}
		if(flag)
			return 1;
	}
	if(m-i>=len)
	{
		p=0;flag=1;
		for(k=i;k<i+len;k++)
			if(s[k][j]!=x[p++])
			{flag=0;break;}
		if(flag)
		return 1;
	}
	if(j+1>=len)
	{
		p=0;flag=1;
		for(k=j;k>=j+1-len;k--)
			if(s[i][k]!=x[p++])
			{flag=0;break;}
		if(flag)
		return 1;
	}
	if(n-j>=len)
	{
		p=0;flag=1;
		for(k=j;k<j+len;k++)//这一开始粗心多打了个=号，wa了5次
			if(s[i][k]!=x[p++])
			{flag=0;break;}
		if(flag)
		return 1;
	}
	if(i+len<=m&&j+len<=n)
	{
		h=i;l=j;flag=1;
		for(k=0;k<len;k++)
			if(x[k]!=s[h++][l++])
				{flag=0;break;}
			if(flag)
			return 1;

	}
	if(i+1>=len&&j+1>=len)
	{
		h=i;l=j;flag=1;
		for(k=0;k<len;k++)
			if(x[k]!=s[h--][l--])
			{flag=0;break;}
		if(flag)
		return 1;
	}
	if(i+1>=len&&j+len<=n)
	{
		h=i;l=j;flag=1;
		for(k=0;k<len;k++)
			if(x[k]!=s[h--][l++])
			{flag=0;break;}
		if(flag)
		return 1;
	}
	if(i+len<=m&&j+1>=len)
	{
		h=i;l=j;flag=1;
		for(k=0;k<len;k++)
			if(x[k]!=s[h++][l--])
			{flag=0;break;}
		if(flag)
		return 1;
	}
	return 0;
}
int main()
{
	int T,i,j,m,n,k,p;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d %d",&m,&n);
		for(i=0;i<m;i++)
		scanf("%s",s[i]);
		for(i=0;i<m;i++)
			for(j=0;j<n;j++)
			s[i][j]=toupper(s[i][j]);
		scanf("%d",&k);
		while(k--)
	    {
		    scanf("%s",x);p=0;
		    for(i=0;i<strlen(x);i++)
				x[i]=toupper(x[i]);
	  	    for(i=0;i<m;i++)
			{
				for(j=0;j<n;j++)
				if(is_find(m,n,i,j))
		        {
		        	printf("%d %d\n",i+1,j+1);
		        	p=1;
		        	break;
		        }
		        if(p)
			    break;
			}
	    }
	    if(T) //每组测试数据空一行 坑点
		printf("\n");
	}
	return 0;
}

UVA 10010-- Where's Waldorf?--暴力串处理