Problem:
Score : 800 points

Problem Statement

Mr. Takahashi has in his room an art object with H rows and W columns, made up of H \times W blocks.
Each block has a color represented by a lowercase English letter (a-z).
The color of the block at the i-th row and j-th column is c_{i,j}.

Mr. Takahashi would like to dismantle the object, finding it a bit kitschy for his tastes.
The dismantling is processed by repeating the following operation:

Choose one of the W columns and push down that column one row.
The block at the bottom of that column disappears.

Each time the operation is performed, a cost is incurred.
Since blocks of the same color have a property to draw each other together, the cost of the operation is the number of the pairs of blocks (p, q) such that:

The block p is in the selected column.

The two blocks p and q are horizontally adjacent (before pushing down the column).

The two blocks p and q have the same color.

Mr. Takahashi dismantles the object by repeating the operation H \times W times to get rid of all the blocks.
Compute the minimum total cost to dismantle the object.

Constraints

1 ≤ H ≤ 300

2 ≤ W ≤ 300

All characters c_{i,j} are lowercase English letters (a-z).

Partial Score

In test cases worth 300 points, W = 3.

Input

The input is given from Standard Input in the following format:

H W
c_{1,1}c_{1,2}..c_{1,W}
c_{2,1}c_{2,2}..c_{2,W}
:
c_{H,1}c_{H,2}..c_{H,W}

Output

Print the minimum total cost to dismantle the object.

Sample Input 1

2 3
rrr
brg

Sample Output 1

2

For example, the total cost of 2 can be achieved by performing the operation as follows and this is the minimum value.

Sample Input 2

6 3
xya
xya
ayz
ayz
xaz
xaz

Sample Output 2

0

The total cost of 0 can be achieved by first pushing down all blocks of the middle column, then all of the left column, and all of the right column.

Sample Input 3

4 2
ay
xa
xy
ay

Sample Output 3

0

The total cost of 0 can be achieved by the following operations:

pushing down the right column one row;

pushing down the left column one row;

pushing down all of the right column;

and pushing down all of the left column.

Sample Input 4

5 5
aaaaa
abbba
ababa
abbba
aaaaa

Sample Output 4

24

Sample Input 5

7 10
xxxxxxxxxx
ccccxxffff
cxxcxxfxxx
cxxxxxffff
cxxcxxfxxx
ccccxxfxxx
xxxxxxxxxx

Sample Output 5

130

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.NoSuchElementException;
public class Main {
	static final long C =  1000000007;
	static final int CY = 1000000000;
	static final int INF = Integer.MAX_VALUE/2;
	StringBuilder sb;
	//セット
	class Set<F,S> {
		F f;
		S s;
		Set(F f, S s) {this.f = f;this.s = s;}
	}
	char[][] c;
	int H;
	int W;
	public void calc() {
		sb = new StringBuilder();


		IO sc = new IO();

		H = sc.nextInt();
		W = sc.nextInt();
		c = new char[H][];
		for (int i=0; i< H; i++) {
			c[i] = sc.next().toCharArray();
		}
		memo = new long[H][H][H];
		for (int i=0; i < H; i++) {
			for (int t=0 ;t < H; t++) {
				Arrays.fill(memo[i][t],-1);
			}
		}
		memo[H-1][H-1][H-1] = 0;

		long ans = msd(0,0,0);



		System.out.println(ans);
	}
	long[][][] memo;
	long msd(int x,int  y,int z) {
		if (x >= H || y >= H || z >= H) return INF;
		if (memo[x][y][z] != -1) {
			return memo[x][y][z];
		}
		long ret = 0;
		long nx = msd(x+1, y,z);
		long ny = msd(x, y+1,z);
		long nz = msd(x,y,z+1);
		for (int i=H-1; i >= 0;i--) {
			if (i + x >= H || i + y >= H) break;
			if (c[i-x][0] == c[i-y][1]) {nx++;
			ny++;
			}
		}

		for (int i=H-1; i >= 0;i--) {
			if ( i + y >= H || i+z >= H) break;
			if (c[i-y][1] == c[i-z][2]) {
				nz++;
				ny++;
			}
		}

		ret = Math.min(nx, Math.min(ny, nz));


		return memo[x][y][z] = ret;
	}

	public static void main(String[] args) {
		Main main = new Main();
		main.calc();

	}
}
class IO extends PrintWriter {
	private final InputStream in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	public IO() { this(System.in);}
	public IO(InputStream source) { super(System.out); this.in = source;}
	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		}else{
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}
	private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
	private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
	private static boolean isNewLine(int c) { return c == '\n' || c == '\r';}
	public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
	public boolean hasNextLine() { while(hasNextByte() && isNewLine(buffer[ptr])) ptr++; return hasNextByte();}
	public String next() {
		if (!hasNext()) {
			throw new NoSuchElementException();
		}
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while(isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	public char[] nextCharArray(int len) {
		if (!hasNext()) {
			throw new NoSuchElementException();
		}
		char[] s = new char[len];
		int i = 0;
		int b = readByte();
		while(isPrintableChar(b)) {
			if (i == len) {
				throw new InputMismatchException();
			}
			s[i++] = (char) b;
			b = readByte();
		}
		return s;
	}
	public String nextLine() {
		if (!hasNextLine()) {
			throw new NoSuchElementException();
		}
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while(!isNewLine(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	public long nextLong() {
		if (!hasNext()) {
			throw new NoSuchElementException();
		}
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while(true){
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			}else if(b == -1 || !isPrintableChar(b)){
				return minus ? -n : n;
			}else{
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) {
			throw new NumberFormatException();
		}
		return (int) nl;
	}
	public char nextChar() {
		if (!hasNext()) {
			throw new NoSuchElementException();
		}
		return (char) readByte();
	}
	public double nextDouble() { return Double.parseDouble(next());}
	public int[] nextIntArray(int n) { int[] a = new int[n]; for(int i=0;i<n;i++) a[i] = nextInt(); return a;}
	public long[] nextLongArray(int n) { long[] a = new long[n]; for(int i=0;i<n;i++) a[i] = nextLong(); return a;}
	public double[] nextDoubleArray(int n) { double[] a = new double[n]; for(int i=0;i<n;i++) a[i] = nextDouble(); return a;}
	public void nextIntArrays(int[]... a) { for(int i=0;i<a[0].length;i++) for(int j=0;j<a.length;j++) a[j][i] = nextInt();}
	public int[][] nextIntMatrix(int n,int m) { int[][] a = new int[n][]; for(int i=0;i<n;i++) a[i] = nextIntArray(m); return a;}
	public char[][] nextCharMap(int n,int m) { char[][] a = new char[n][]; for(int i=0;i<n;i++) a[i] = nextCharArray(m); return a;}
	public void close() { super.close(); try {in.close();} catch (IOException e) {}}
}