UVA 10010 - Where's Waldorf?

题目大意：在字符串图里，找出名字的位置，名字可以往八个方向找。不区分大小写。

解题思路：方向数组来移动，先将所有字符串改为小写，输入名字也改为小写。在字符串图中查到名字首字符就进去函数，往八个方向尝试能否得到名字。

ac代码：

#include <iostream>
#include <cstring>
using namespace std;
int dx[8]={-1,1,0,0,-1,-1,1,1}, dy[8]={0,0,-1,1,-1,1,-1,1};
int n, x, y, r, l, k, len;
char puz[105][105], name[105], temp[105];
bool solve(int a, int b)
{
int t1, t2;
for (int i=0; i<8; i++){
t1 = a, t2 = b;
memset(temp, 0, sizeof(temp));
for (int j=0; j<len; j++){
temp[j] = puz[t1][t2];
t1 += dx[i], t2 += dy[i];
}
if (!strcmp(temp, name)){
x = a + 1, y = b + 1;
return 1;
}
}
return 0;
}
int main()
{
scanf("%d", &n);
while (n--){
scanf("%d%d", &r, &l);
for (int i=0; i<r; i++){
scanf("%s", puz[i]);
for (int j=0; j<l; j++)
if (puz[i][j] >= 'a' && puz[i][j] <= 'z')
puz[i][j] = 'A' + puz[i][j] - 'a';
}
scanf("%d", &k);
for (int i=0; i<k; i++){
scanf("%s", name);
x = y = -1;
len = strlen(name);
for (int j=0; j<len; j++)
if (name[j] >= 'a' && name[j] <= 'z')
name[j] = 'A' + name[j] - 'a';
for (int i=0; i<r; i++){
for (int j=0; j<l; j++)
if (puz[i][j] == name[0] && solve(i, j))
break;
if (x != -1)
break;
}
printf("%d %d\n", x, y);
}
if (n)
printf("\n");
}
return 0;
}