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
算是寒假后续刷题那套的后续吧。。。
输入量太大调试令人蛋疼，今后得用些方法解决调试问题，现在时间急不想搞。。。
利用一个特长的函数来判断位置非常郁闷，所谓懒婆娘的裹脚布，调试了一下午终于正常了。
代码如下（非Ac）：
#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;
}
这样看来的确又臭又长。。。
而且以上代码虽然结果和例子一样，但却ac不了。。。
后来找刷过的朋友问了下原来里面有每组之间空行的条件。这种错误是第几次了。。。
用if判断一下，还是ac不了。
后来那别人的代码来测试对比，发现逻辑有错误，我原来的做法是：找到字符后，用行列的加减判断是否可以按那个方向排字符串，再录入字符串用strcmp进行判断，因为我觉得letter[-1][-3]这种数组会出错（证实不会，人家会空着）。。。
结果判断时出错了，搞了好久都没弄好，最后干脆放弃运行效率（虽然本来我那程序效率就不高），找到字符就各个方向都判断不进行筛选，最后，终于AC了！！！搞了好几天。。。
贴AC代码（略长，都是重复的）：
#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
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;
}
这题很水，是水中比较有难度的，但我却折腾那么久，我太弱了，入手本本后一定要多练了。。。
