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 算是寒假後續刷題那套的後續吧。。。 輸入量太大調試令人蛋疼，今後得用些方法解決調試問題，現在時間急不想搞。。。 利用一個特長的函數來判斷位置非常郁悶，所謂懶婆娘的裹腳布，調試了一下午終於正常了。 代碼如下（非Ac）： [cpp] #include<stdio.h> #include<string.h> #include<ctype.h> int i, j, l,cnt, n, m, k, rec_m, rec_n; char letter[50][50], word[20][50], temp[50]; int search(void); int main() { scanf("%d", &cnt); while (cnt --) { scanf("%d%d", &m, &n); for (i = 0; i < m; i ++) { scanf("%s", letter[i]); for (j = 0; j < strlen(letter[i]); j ++) letter[i][j] = tolower(letter[i][j]); } scanf("%d", &k); for (i = 0; i < k; i ++) { scanf("%s", word[i]); for (j = 0; j < strlen(word[i]); j ++) word[i][j] = tolower(word[i][j]); } for (i = 0; i < k; i ++) { search(); printf("%d %d\n", rec_n + 1, rec_m + 1); } } return 0; } int search(void) { int len = strlen(word[i]); for (j = 0; j < m; j ++) { for (l = 0; l < n; l ++) { if (word[i][0] == letter[j][l]) { if (l + len <= n) { if (j - len + 1 >= 0) { //ru memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j + p][l - p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } } if (j + len <= m) { //rd memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j + p][l + p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } } //r memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j + p][l]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } } if (l - len + 1 >= 0) { if (j - len + 1 >= 0) { //lu memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j - p][l - p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } } if (j + len <= m) { //ld memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j - p][l + p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } } //l memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j - p][l]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } } if (j - len + 1 >= 0) { //u memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j][l - p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } } if (j + len <= m) { //d memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j][l + p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } } } } } rec_m = rec_n = 0; return 0; } 這樣看來的確又臭又長。。。 而且以上代碼雖然結果和例子一樣，但卻ac不了。。。 後來找刷過的朋友問了下原來裡面有每組之間空行的條件。這種錯誤是第幾次了。。。 用if判斷一下，還是ac不了。 後來那別人的代碼來測試對比，發現邏輯有錯誤，我原來的做法是：找到字符後，用行列的加減判斷是否可以按那個方向排字符串，再錄入字符串用strcmp進行判斷，因為我覺得letter[-1][-3]這種數組會出錯（證實不會，人家會空著）。。。 結果判斷時出錯了，搞了好久都沒弄好，最後干脆放棄運行效率（雖然本來我那程序效率就不高），找到字符就各個方向都判斷不進行篩選，最後，終於AC了！！！搞了好幾天。。。 貼AC代碼（略長，都是重復的）： [html] #include<stdio.h> #include<string.h> #include<ctype.h> int i, j, l,cnt, n, m, k, rec_m, rec_n; char letter[50][50], word[20][50], temp[50]; int search(void); int main() { scanf("%d", &cnt); while (cnt --) { scanf("%d%d", &m, &n); for (i = 0; i < m; i ++) { scanf("%s", letter[i]); for (j = 0; j < strlen(letter[i]); j ++) letter[i][j] = tolower(letter[i][j]); } scanf("%d", &k); for (i = 0; i < k; i ++) { scanf("%s", word[i]); for (j = 0; j < strlen(word[i]); j ++) word[i][j] = tolower(word[i][j]); } for (i = 0; i < k; i ++) { search(); printf("%d %d\n", rec_n + 1, rec_m + 1); } if (cnt != 0) printf(("\n")); } return 0; } int search(void) { int len = strlen(word[i]); for (j = 0; j < m; j ++) { for (l = 0; l < n; l ++) { if (word[i][0] == letter[j][l]) { //ru memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j + p][l - p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } //rd memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j + p][l + p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } //r memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j + p][l]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } //lu memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j - p][l - p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } //ld memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j - p][l + p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } //l memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j - p][l]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } //u memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j][l - p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } //d www.2cto.com memset(temp, 0, sizeof(temp)); for (int p = 0; p < len; p++) temp[p] = letter[j][l + p]; if (strcmp(temp, word[i]) == 0) { rec_n = j; rec_m = l; return 0; } } } } rec_m = rec_n = 0; return 0; }
