E - GirlCat

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×mn×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 TT which represents the case number.

As for each case, the first line are two integers nn and mm, which are the height and the width of the photo.
Then there are nn lines followed, and there are mm characters of each line, which are the the details of the photo.

It is guaranteed that:
TT is about 50.
1≤n≤10001≤n≤1000.
1≤m≤10001≤m≤1000.
∑(n×m)≤2×106∑(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

題意：給你一張含有圖，找cat 和 girl 有多少個（只要相鄰就行 而且有一個字母沒用過就算一個）
方法：dfs

#include <stdio.h>
#include <string.h>
char a[1004][1004];
int pos[1000004];
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
int sumcat,sumgirl;
int n,m;
int judge2(int x, int y, int tempx, int tempy)
{
    if(a[x][y] == 'g'&&a[tempx][tempy] =='i') return 1;
    if(a[x][y] == 'i'&&a[tempx][tempy] =='r') return 1;
    if(a[x][y] == 'c'&&a[tempx][tempy] =='a') return 1;
    return 0;
}
int judge(int x,int y,int tempx,int tempy)
{
    if((tempx>=0&&tempx<n&&tempy>=0&&tempy<m))
       {
           if(judge2(x,y,tempx,tempy))
           return 1;
           else return 0;
       }
    else return 0;
}
void dfs(int num)
{
    int x = num / m;
    int y = num % m;
    for(int i = 0;i < 4; i++)
    {
        int tempx = x + dx[i];
        int tempy = y + dy[i];
        if(tempx>=0&&tempx<n&&tempy>=0&&tempy<m&&a[x][y]=='a'&&a[tempx][tempy] == 't')
        {
            sumcat++;
            continue;
        }
        if(tempx>=0&&tempx<n&&tempy>=0&&tempy<m&&a[x][y]=='r'&&a[tempx][tempy] == 'l')
        {
            sumgirl++;
            continue;
        }
        if(judge(x,y,tempx,tempy))
        {
            dfs(tempx*m+tempy);
        }
    }
    return ;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(pos,0,sizeof(pos));
        scanf("%d%d",&n,&m);
        sumgirl = 0;
        sumcat = 0;
        int p = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                scanf(" %c",&a[i][j]);
                if(a[i][j] == 'c' || a[i][j] == 'g')
                {
                    pos[p++] = i*m + j;
                }
            }
        }
        for(int i = 0; i < p; i++)
        {
            dfs(pos[i]);
        }
        printf("%d %d\n",sumgirl,sumcat);
    }
}