Looksery Cup 2015 A. Face Detection 水题

A. Face Detection

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/549/problem/A

Description

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 Input

4 4
xxxx
xfax
xcex
xxxx

Sample Output

1

HINT

题意

 给你n*m的格子，格子里面有字符串，然后问你有多少个2*2的格子里面，含有face

题解：

随便产生一组全排列，然后枚举就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 2000001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

string s[100];
string s1;
int c[4]={0,1,2,3};
vector<int> a[50];
void unit()
{
    int i = 0;
    do
    {
        for(int j = 0; j < 4; j++)
            a[i].push_back(c[j]);
        i++;
    } while(next_permutation(c,c + 4));
}
int main()
{
    //test;
    unit;
    int n=read(),m=read();
    for(int i=0;i<n;i++)
        cin>>s[i];
    s1="face";
    int ans=0;
    int c[4]={0,1,2,3};
    for(int i=0;i<n-1;i++)
    {
        for(int j=0;j<m-1;j++)
        {
            c[0]=0,c[1]=1,c[2]=2,c[3]=3;
            do{
                if(s[i][j]==s1[c[0]]&&s[i+1][j]==s1[c[1]]&&s[i+1][j+1]==s1[c[2]]&&s[i][j+1]==s1[c[3]])
                {
                    ans++;
                    break;
                }
            }while(next_permutation(c,c+4));
        }
    }
    cout<<ans<<endl;
}    

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.