GirlCat

GirlCat

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1909    Accepted Submission(s): 1128

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 playing ``Hide 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' to `z').
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 exactly ``cat'' 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

思路：找到g和c开头的dfs就好了。

#include<bits/stdc++.h>
#define REP(i, a, b) for(int i = (a); i <= (b); ++ i)
#define REP(j, a, b) for(int j = (a); j <= (b); ++ j)
#define PER(i, a, b) for(int i = (a); i >= (b); -- i)
using namespace std;
const int maxn=1e5+5;
template <class T>
inline void rd(T &ret){
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9'){
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}
char str[5][10]={"remove","girl","cat"};
char p[1005][1005];
int T,q[5],m,n;
void dfs(int x,int y,int cnt,int cur){
      if(x<1||y<1||x>n||y>m||p[x][y]!=str[cur][cnt])return;
      if((cur==1&&cnt==3)||(cur==2&&cnt==2)){
             q[cur]++;
             return;
      }
      dfs(x+1,y,cnt+1,cur);
      dfs(x,y+1,cnt+1,cur);
      dfs(x-1,y,cnt+1,cur);
      dfs(x,y-1,cnt+1,cur);
}
int main()
{
    rd(T);
    while(T--){
        rd(n),rd(m);
        memset(q,0,sizeof(q));
        for(int i=1;i<=n;i++)scanf("%s",p[i]+1);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(p[i][j]=='g')dfs(i,j,0,1);
                if(p[i][j]=='c')dfs(i,j,0,2);
            }
        }
        cout<<q[1]<<' '<<q[2]<<endl;
    }
    return 0;
}