Face Detection



Face Detection

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

给一个矩阵的字符串，判断是否存在一个2*2矩形，里面有face四个字母。第一想法是搜索，然后枚举四个方向，最后除以4。

关键在于判断，第一次没想到简单方法，后来采用leap标记的方法，c[i][j]-'a'来表示第几个字母，在判断face对应的是否是访问过，就可以ans++了。

当时的leap开小了，数组越界了，wa的我莫名其妙，最近好几题都是数组大小问题。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
char c[100][100];
int ans, leap[30];
void bfs(int x,int y)
{
	leap[0] = leap[2] = leap[5] = leap[4] = 0;
	leap[c[x][y] - 'a'] = 1; leap[c[x + 1][y] - 'a'] = 1; leap[c[x + 1][y + 1] - 'a'] = 1; leap[c[x][y + 1] - 'a'] = 1;
	if (leap[0] && leap[2] && leap[5] && leap[4])ans++;

	leap[0] = leap[2] = leap[5] = leap[4] = 0;
	leap[c[x][y] - 'a'] = 1; leap[c[x][y + 1] - 'a'] = 1; leap[c[x - 1][y] - 'a'] = 1; leap[c[x - 1][y + 1] - 'a'] = 1;
	if (leap[0] && leap[2] && leap[5] && leap[4])ans++;

	leap[0] = leap[2] = leap[5] = leap[4] = 0;
	leap[c[x][y] - 'a'] = 1; leap[c[x + 1][y] - 'a'] = 1; leap[c[x + 1][y - 1] - 'a'] = 1; leap[c[x][y - 1] - 'a'] = 1;
	if (leap[0] && leap[2] && leap[5] && leap[4])ans++;

	leap[0] = leap[2] = leap[5] = leap[4] = 0;
	leap[c[x][y] - 'a'] = 1; leap[c[x][y - 1] - 'a'] = 1; leap[c[x - 1][y] - 'a'] = 1; leap[c[x - 1][y - 1] - 'a'] = 1;
	if (leap[0] && leap[2] && leap[5] && leap[4])ans++;
}
int main()
{
	int i, j, m, n;
	cin >> n >> m;
	getchar();
	for (i = 0; i <= n + 1; i++)
	for (j = 0; j <= m + 1; j++)
		c[i][j] = 'x';
	for (i = 1; i <= n; i++)
	{
		for (j = 1; j <= m; j++)
		{
			scanf("%c", &c[i][j]);
		}
		getchar();
	}
	ans = 0;
	for (i = 1; i <= n; i++)
	{
		for (j = 1; j <= m; j++)
		{
			if (c[i][j] == 'a'||c[i][j]=='f'||c[i][j]=='c'||c[i][j]=='e')
				bfs(i, j);
		}
	}
	cout << ans / 4 << endl;
	return 0;
}

2.其实现在想想没有必要4个方向都搜索，可以只搜索c[x][y],c[x+1][y],c[x][y+1],c[x+1][y+1],因为只要存在，就一定可以通过左上角的把他判断出来。贴上tourist的代码，好好学习，还是基础太差，否则完全可以用string和sort熟练搞定。

#include <bits/stdc++.h>

using namespace std;

char s[777][777];

int main() {
  int h, w;
  scanf("%d %d", &h, &w);
  for (int i = 0; i < h; i++) {
    scanf("%s", s[i]);
  }
  int ans = 0;
  for (int i = 0; i < h - 1; i++) {
    for (int j = 0; j < w - 1; j++) {
      string z = "";
      z += s[i][j];
      z += s[i][j + 1];
      z += s[i + 1][j];
      z += s[i + 1][j + 1];
      sort(z.begin(), z.end());
      if (z == "acef") {
        ans++;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}