hdoj-5706-GirlCat

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

简单的dfs，要求连续的cat和girl的个数，然后四个方向搜一搜

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#define exp 1e-10using namespace std;const int N = 1005;const int M = 15;const int inf = 100000000;const int mod = 2009;char s[N][N],ch[2][5]={"girl","cat"};bool v[N][N];int ans[2],n,m;bool check(int x,int y){    return x>=0&&x<n&&y>=0&&y<m;}void dfs(int x,int y,int k,int t){    if(!check(x,y)||s[x][y]!=ch[t][k])        return;    v[x][y]=true;    if(t&&k==2||!t&&k==3)    {        ans[t]++;        return;    }    dfs(x+1,y,k+1,t);    dfs(x,y+1,k+1,t);    dfs(x-1,y,k+1,t);    dfs(x,y-1,k+1,t);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        ans[0]=ans[1]=0;        memset(v,false,sizeof(v));        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(!v[i][j])                {                    if(s[i][j]=='g')                        dfs(i,j,0,0);                    else if(s[i][j]=='c')                        dfs(i,j,0,1);                }        printf("%d %d\n",ans[0],ans[1]);    }    return 0;}