1014: The Matrix

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本文介绍了一个基于字符矩阵寻找指定单词的问题。文章详细解释了输入输出格式，并提供了一段C语言代码示例，演示如何判断单词是否存在于给定的字符矩阵中。

1014: The Matrix

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	8192K	1888	572	Standard

Given a matrix of characters, and a list of words, output whether or not each word is present in the matrix. Words may appear forwards and backwards. They may appear horizontally, vertically, and diagonally.

Input

The matrix, followed by a list of words. Only lower case characters will be used. Each matrix will be square, and contain no more than 20 characters on a side. The string XXX denotes the end of input.

Output

For each word, report whether or not it is present in the matrix. If it is present, output should read “<word> is in the matrix.” If it is not present, output should read “<word> is not in the matrix.”, where <word> is the word in question.

Sample Input

applexy
pxlhjke
edeqgfl
xocgvpl
gghnmnn
tahuupu
qdgbywb
apple
axe
apex
cat
car
hat
computer
gum
XXX

Sample Output

apple is in the matrix.
axe is in the matrix.
apex is in the matrix.
cat is not in the matrix.
car is not in the matrix.
hat is in the matrix.
computer is not in the matrix.
gum is in the matrix.

#include<stdio.h>
#include<string.h>
int main()
{
 char s[30][30];
 scanf("%s",s[0]);
 int len=strlen(s[0]);
 int i;
 for(i=1;i<len;i++)
  scanf("%s",s[i]);
 char c[30];
 //8个方向
 while(scanf("%s",c)!=-1)
 {
  if(strcmp(c,"XXX")==0) break;
  printf("%s",c);
  int f=0,k=0;
  int i,j;
  for(i=0;i<len;i++)
  {
   for(j=0;j<len;j++)
   {
    if(s[i][j]==c[0])
    {
     for(k=0;k<strlen(c);k++)
     {
      //上(i>k)
      if(i<k) break;
      if(c[k]!=s[i-k][j]) break;
      if(k==strlen(c)-1)  f=1;
     }
     for(k=0;k<strlen(c);k++)
     {
      //右上
      if(i<k||j+k>=len) break;
      if(c[k]!=s[i-k][j+k]) break;
      if(k==strlen(c)-1) f=1;
     }
     for(k=0;k<strlen(c);k++)
     {
      //右
      if(j+k>=len) break;
      if(c[k]!=s[i][j+k]) break;
      if(k==strlen(c)-1) f=1;
     }
     for(k=0;k<strlen(c);k++)
     {
      //右下
      if(j+k>=len||i+k>=len) break;
      if(c[k]!=s[i+k][j+k]) break;
      if(k==strlen(c)-1) f=1;
     }
     for(k=0;k<strlen(c);k++)
     {
      //下
      if(i+k>=len) break;
      if(c[k]!=s[i+k][j]) break;
      if(k==strlen(c)-1) f=1;
     }
     for(k=0;k<strlen(c);k++)
     {
      //左下
      if(j<k||i+k>=len) break;
      if(c[k]!=s[i+k][j-k]) break;
      if(k==strlen(c)-1) f=1;
     }
     for(k=0;k<strlen(c);k++)
     {
      //左
      if(j<k) break;
      if(c[k]!=s[i][j-k]) break;
      if(k==strlen(c)-1) f=1;
     }
     for(k=0;k<strlen(c);k++)
     {
      //左上
      if(j<k||i<k) break;
      if(c[k]!=s[i-k][j-k]) break;
      if(k==strlen(c)-1) f=1;
     }
    }
   }
  }
  if(f) 
   printf(" is in the matrix./n");
  else 
   printf(" is not in the matrix./n");
 }
 return 0;
}