Looksery Cup 2015 A. Face Detection

怎么说呢，这个题还是很简单的，循环遍历矩阵，每个位置上的4个数[ i ][ j ]  ,[ i + 1][ j ], [ i ][ j + 1], [i + 1][j + 1].对应的字符+1，判断是不是f，a，c，e。如果是cnt++，然后ok清零

如果不是ok直接清零，防止影响后边的计算。

The developers of Looksery have to write an efficient algorithm that detects faces on a picture. Unfortunately, they are currently busy preparing a contest for you, so you will have to do it for them.

In this problem an image is a rectangular table that consists of lowercase Latin letters. A face on the image is a 2 × 2 square, such that from the four letters of this square you can make word "face".

You need to write a program that determines the number of faces on the image. The squares that correspond to the faces can overlap.

Input

The first line contains two space-separated integers, n and m (1 ≤ n, m ≤ 50) — the height and the width of the image, respectively.

Next n lines define the image. Each line contains m lowercase Latin letters.

Output

In the single line print the number of faces on the image.

Sample test(s)

Input

4 4
xxxx
xfax
xcex
xxxx

Output

1

Input

4 2
xx
cf
ae
xx

Output

1

Input

2 3
fac
cef

Output

2

Input

1 4
face

Output

0

Note

In the first sample the image contains a single face, located in a square with the upper left corner at the second line and the second column:

In the second sample the image also contains exactly one face, its upper left corner is at the second row and the first column.

In the third sample two faces are shown:

In the fourth sample the image has no faces on it.

#include<cstdio>
#include<cmath>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;

char a[55][55];

int main()
{
    //freopen("1.txt", "r", stdin);
    int ok[200]={0};
    int cnt = 0;
    int x ,y;
    cin>>x>>y;
    getchar();
    for(int i = 0; i < x; i++)
        gets(a[i]);
//    for(int i = 0; i < x; i++)
//        puts(a[i]);
    for(int i = 0; i < x; i++)
    {
        for(int j = 0 ; j < y ; j++)
        {
            if( i >= 0 && i + 1 < x && j >= 0 && j + 1 < y )
            {
                //cout<<i<<" "<<j<<endl;
                ok[a[i][j]]++;
               // cout <<ok[a[i][j]] << " " <<a[i][j]  <<endl;
                ok[a[i+1][j]]++;
                ok[a[i][j+1]]++;
                ok[a[i+1][j+1]]++;
                if(ok['f'] != 0 && ok['a'] != 0 && ok['c'] != 0 && ok['e'] != 0)
                {
                    cnt++;
                    memset(ok, 0, sizeof(ok));
                }
                else
                {
                    memset(ok , 0 , sizeof(ok));
                }

            }
        }
    }
    cout<<cnt<<endl;

    return 0;
}