用java语言实现以下算法，算法描述如下：小红有一个大小为 n × m 的棋盘，”."表示这个格子没有棋子，"X”表示这个格子有棋子。第i行第j列的格子可以用一个坐标（i,j）表示

以下是用Java语言实现该算法的代码：

```java
import java.util.Scanner;

public class SquareCount {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
scanner.nextLine(); // 读取换行符

char[][] chessboard = new char[n][m];
for (int i = 0; i < n; i++) {
String line = scanner.nextLine();
for (int j = 0; j < m; j++) {
chessboard[i][j] = line.charAt(j);
}
}

int count = countSquares(chessboard);
System.out.println(count);
}

private static int countSquares(char[][] chessboard) {
int n = chessboard.length;
int m = chessboard[0].length;
int count = 0;

// 遍历所有可能的正方形顶点
for (int x1 = 0; x1 < n; x1++) {
for (int y1 = 0; y1 < m; y1++) {
if (chessboard[x1][y1] == 'X') {
for (int x2 = 0; x2 < n; x2++) {
for (int y2 = 0; y2 < m; y2++) {
if (chessboard[x2][y2] == 'X') {
int dx = x2 - x1;
int dy = y2 - y1;
int x3 = x2 + dy;
int y3 = y2 - dx;
int x4 = x1 + dy;
int y4 = y1 - dx;

if (isValidPoint(x3, y3, n, m) && isValidPoint(x4, y4, n, m)
&& chessboard[x3][y3] == 'X' && chessboard[x4][y4] == 'X') {
count++;
}
}
}
}
}
}
}

return count;
}

private static boolean isValidPoint(int x, int y, int n, int m) {
return x >= 0 && x < n && y >= 0 && y < m;
}
}
```

使用该代码，你可以输入棋盘的大小和棋盘的布局，然后计算出小红有多少种选取四个棋子构成正方形的方案。