Zju1290 Word-Search Wonder(http://begin.lydsy.com/JudgeOnline/problem.php?id=2768)

2768: Zju1290 Word-Search Wonder

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 4  Solved: 2
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Description

The Pyrates Restaurant was starting to fill up as Valentine McKee walked in. She scanned the crowd for her sister, brother-in-law, and nephew. Seeing her sister waving from the far end of the restaurant, she made her way back to their booth. ``Hi, Valentine,'' her sister and brother-in-law, Niki and Dennis Chapman, greeted her.
``Hi, guys,'' she replied. ``What are you doing, Wade?'' she asked her nephew. He was busy working on one of the restaurant's activity sheets with a crayon.
``I'm doing a word search game,'' Wade explained. ``I have to find all of these words in this big mess of letters. This is really hard.'' Wade looked intently at the paper in front of him.
``Can I help?'' asked Valentine, looking across the table at the activity sheet.
``Sure. These are the words we're looking for. They're the names of different kinds of Planes, Trains, and Automobiles.''
在字母矩阵找找单词游戏，找的方向有8个，水平、垂直、两个对角线，外加每种两个方向。求给定的词是否在字母矩阵中，并求开始和结束的坐标。

Input

The first line of input will specify the length (in characters) of the sides of the letter matrix (the matrix of letters will be square). The length, l, will be in the range 1 <= l <= 100. The next l lines of input will be the matrix itself, each line will contain l uppercase letters.

A list of words will follow. Each word will be on a line by itself; there will be 100 or fewer words. Each word will be 100 or fewer characters long, and will only contain uppercase letters.

The final line of input will contain a single zero character.

Output

Your program should attempt to find each word from the word list in the puzzle. A word is ``found'' if all the characters in the word can be traced in a single (unidirectional) horizontal, vertical, or diagonal line in the letter matrix. Words may not ``wrap around'' rows or columns, but horizontal and diagonal words may proceed from right to left (``backwards''). For each word that is found, your program should print the coordinates of its first and last letters in the matrix on a single line, separated by a single space. Coordinates are pairs of comma-separated integers (indexed from 1), where the first integer specifies the row number and the second integer specifies the column number.

If a word is not found, the string ``Not found'' should be output instead of a pair of coordinates.

Each word from the input can be ``found'' at most once in the puzzle.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

1

5
EDEEE
DISKE
ESEEE
ECEEE
EEEEE
DISC
DISK
DISP
0

Sample Output

1,2 4,2
2,1 2,4
Not found

HINT

 

Source

Trie系列

题解：

　　分类是trie树，写了一个dfs过了，醉！

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 using namespace std;
 char a[][],s[];
 int n,m,i,j;
 int x1,y1,x2,y2;
 bool dfs(int deep,int x,int y,int dx,int dy)
 {
     if (deep>strlen(s+)){x2=x-dx; y2=y-dy; return true;}
     if (x>n || x<= || y>n|| y<=) return false;
     if (a[x][y]!=s[deep]) return false;
     return dfs(deep+,x+dx,y+dy,dx,dy);
 }
 bool find (char *s)
 {
     for (int i=; i<=n; i++)
         for (int j=; j<=n; j++)
             if (a[i][j]==s[])
             {
                 for (int k=-; k<=; k++)
                     for (int kk=-; kk<=; kk++)
                     if (k!= || kk!=)
                     if (dfs(,i,j,k,kk))
                     {
                         x1=i; y1=j; return true;
                     }
             }
     return false;
 }
 void work()
 {
     cin>>n;
     for (int i=; i<=n; i++) for (int j=; j<=n; j++) cin>>a[i][j];
     while (true)
     {
         scanf("%s",s+);
         if (s[]=='') break;
         if (find(s)) cout<<x1<<','<<y1<<' '<<x2<<','<<y2<<endl; else cout<<"Not found"<<endl;
     }
 }
 int main()
 {
     work();
 }