题目链接：acm.hdu.edu.cn/showproblem.php?pid=5706
GirlCat
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1823 Accepted Submission(s): 1091
Problem Description
As a cute girl, Kotori likes playing Hide and Seek'' with cats particularly. Under the influence of Kotori, many girls and cats are playingHide and Seek’’ together.
Koroti shots a photo. The size of this photo is n×m, each pixel of the photo is a character of the lowercase(from a' toz’).
Kotori wants to know how many girls and how many cats are there in the photo.
We define a girl as – we choose a point as the start, passing by 4 different connected points continuously, and the four characters are exactly girl'' in the order. We define two girls are different if there is at least a point of the two girls are different. We define a cat as -- we choose a point as the start, passing by 3 different connected points continuously, and the three characters are exactlycat’’ in the order.
We define two cats are different if there is at least a point of the two cats are different.
Two points are regarded to be connected if and only if they share a common edge.
Input
The first line is an integer T which represents the case number.
As for each case, the first line are two integers n and m, which are the height and the width of the photo.
Then there are n lines followed, and there are m characters of each line, which are the the details of the photo.
It is guaranteed that:
T is about 50.
1≤n≤1000.
1≤m≤1000.
∑(n×m)≤2×106.
Output
As for each case, you need to output a single line.
There should be 2 integers in the line with a blank between them representing the number of girls and cats respectively.
Please make sure that there is no extra blank.
Sample Input
3
1 4
girl
2 3
oto
cat
3 4
girl
hrlt
hlca
Sample Output
1 0
0 2
4 1
题目大意：输入第一行表示多少个测试案例，然后跟着T组测试案例，每一组都先输入多少行，多少列。然后紧跟着就是n行m列字符。然后让你找这个图中有多少个girl和cat,比如girl你就从g字符开始从它的上下左右找i,找到i后，又从i的上下左右找r。一次如果能组成一个完整的girl就算一个。cat也是这样。
解题思路：
典型的dfs,bfs亦可。这里用的是dfs,用两个字符数组将girl和cat存起来，需要注意的是每次递归传值时，都要把字符数组的下标传进去，遍历四个方向时利用传进来的下标取字符串中的字符来比较。思路虽然简单，但还是要注意dfs的细节。
AC代码：
#include<cstdio>
int M[4][2] = {-1, 0, 1, 0, 0, 1, 0, -1};
char s1[] = "girl", s2[] = "cat";
char s[1010][1010];
int ans1, ans2;
int n, m;
void dfsgirl(int x, int y, int k)
{
int dx, dy;
for(int i = 0; i < 4; ++i)
{
dx = x + M[i][0];
dy = y + M[i][1];
if(dx >= 0 && dx < n && dy >= 0 && dy < m && s[dx][dy] == s1[k])
{
k++;
if(k == 4)
{
ans1++;
}
dfsgirl(dx, dy, k);
k--;
}
}
}
void dfscat(int x, int y, int k)
{
int dx, dy;
for(int i = 0; i < 4; ++i)
{
dx = x + M[i][0];
dy = y + M[i][1];
if(dx >= 0 && dx < n && dy >= 0 && dy < m && s[dx][dy] == s2[k])
{
k++;
if(k == 3)
{
ans2++;
k--;
}
else
{
dfscat(dx, dy, k);
k--;
}
}
}
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
ans1 = ans2 = 0;
scanf("%d%d", &n, &m);
for(int i = 0; i < n; ++i)
scanf("%s", s[i]);
for(int i = 0; i < n; ++i)
for(int j = 0; j < m; ++j)
{
if(s[i][j] == 'g')
{
dfsgirl(i, j, 1);
}
if(s[i][j] == 'c')
{
dfscat(i, j, 1);
}
}
printf("%d %d\n", ans1, ans2);
}
return 0;
}