UVA 題目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 

18 11abcDEFGhigghEbkWalDorkFtyAwaldORmFtsimrLqsrcbyoArBeDeyvKlcbqwikomkstrEBGadhrbyUiqlxcnBjf4WaldorfBambiBettyDagbert

Sample Output 

2 52 31 27 8

Miguel Revilla 
2000-08-22

【大意】：

輸入：

給你一個由字母組成的網格，M行N列。尋找一個單詞在網格中的位置。一個單詞匹配網格中聯絡不間斷的字母。可以沿任意方向匹配，一共可以匹配八個方向。忽略大小寫。

需要匹配的字串有K個。

輸出：

每組輸出之間都一行空行。

m n：m代表匹配的最上面的行

      n代表匹配的最下面的行

如果結果有多個，只輸出匹配串。要求：匹配串的第一個字母必須是最高最左的。結果至少有一個。

【代碼】：

/**********************************   日期：2013-4-23*   作者：SJF0115*   題號: 題目10010 - Where's Waldorf?*   來源：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=12&page=show_problem&problem=951*   結果：AC*   來源：UVA*   總結：**********************************/#include<stdio.h>#include<string.h>char Matrix[51][51];char str[21],temp[21];int StartR,StartC;//M行 N列int Match(int M,int N,int &StartR,int &StartC){int i,j,k,flag;StartR = 51,StartC = 51;int len = strlen(str);for(i = 0;i < M;i++){for(j = 0;j < N;j++){flag = 1;//left - rightif(j + len <= N){flag = 0;for(k = 0;k < len;k++){if(str[k] != Matrix[i][j+k]){flag = 1;break;}}if(flag == 0){if(StartR >  i+1){StartR = i+1;StartC = j+1;}else if(StartR ==  i+1 && StartC > j+1){StartR = i+1;StartC = j+1;}}}//right - leftif(j - len + 1>= 0){flag = 0;for(k = 0;k < len;k++){if(str[k] != Matrix[i][j-k]){flag = 1;break;}}if(flag == 0){if(StartR >  i+1){StartR = i+1;StartC = j+1;}else if(StartR ==  i+1 && StartC > j+1){StartR = i+1;StartC = j+1;}}}//up - downif(i + len <= M){flag = 0;for(k = 0;k < len;k++){if(str[k] != Matrix[i+k][j]){flag = 1;break;}}if(flag == 0){if(StartR >  i+1){StartR = i+1;StartC = j+1;}else if(StartR ==  i+1 && StartC > j+1){StartR = i+1;StartC = j+1;}}}//down - upif(i - len + 1 >= 0){flag = 0;for(k = 0;k < len;k++){if(str[k] != Matrix[i-k][j]){flag = 1;break;}}if(flag == 0){if(StartR >  i+1){StartR = i+1;StartC = j+1;}else if(StartR ==  i+1 && StartC > j+1){StartR = i+1;StartC = j+1;}}}//right - upif(j + len <= N && i - len + 1 >= 0){flag = 0;for(k = 0;k < len;k++){if(str[k] != Matrix[i-k][j+k]){flag = 1;break;}}if(flag == 0){if(StartR >  i+1){StartR = i+1;StartC = j+1;}else if(StartR ==  i+1 && StartC > j+1){StartR = i+1;StartC = j+1;}}}//right - downif(j + len <= N && i + len <= M){flag = 0;for(k = 0;k < len;k++){if(str[k] != Matrix[i+k][j+k]){flag = 1;break;}}if(flag == 0){if(StartR >  i+1){StartR = i+1;StartC = j+1;}else if(StartR ==  i+1 && StartC > j+1){StartR = i+1;StartC = j+1;}}}//left - upif(j - len + 1 >= 0 && i - len + 1 >= 0){flag = 0;for(k = 0;k < len;k++){if(str[k] != Matrix[i-k][j-k]){flag = 1;break;}}if(flag == 0){if(StartR >  i+1){StartR = i+1;StartC = j+1;}else if(StartR ==  i+1 && StartC > j+1){StartR = i+1;StartC = j+1;}}}//left - downif(j - len + 1 >= 0 && i + len <= M){flag = 0;for(k = 0;k < len;k++){if(str[k] != Matrix[i-k][j+k]){flag = 1;break;}}if(flag == 0){if(StartR >  i+1){StartR = i+1;StartC = j+1;}else if(StartR ==  i+1 && StartC > j+1){StartR = i+1;StartC = j+1;}}}}//for j}//for ireturn 0;}int main (){int i,j,Case,k,M,N;//freopen("C:\\Users\\XIAOSI\\Desktop\\acm.txt","r",stdin);  while(scanf("%d",&Case) != EOF){while(Case--){scanf("%d %d",&M,&N);//輸入字元矩陣for(i = 0;i < M;i++){scanf("%s",temp);for(j = 0;j < N;j++){Matrix[i][j] = temp[j];//轉換為小寫if(Matrix[i][j] >= 'A' && Matrix[i][j] <= 'Z'){Matrix[i][j] = Matrix[i][j] - 'A' + 'a';}}}scanf("%d",&k);//待匹配串for(i = 0;i < k;i++){scanf("%s",str);int len = strlen(str);//轉換為小寫for(j = 0;j < len;j++){if(str[j] >= 'A' && str[j] <= 'Z'){str[j] = str[j] - 'A' + 'a';}}//printf("%s",str);Match(M,N,StartR,StartC);printf("%d %d\n",StartR,StartC);}//每組測試之間有空行if(Case){printf("\n");}}}return 0;}