Problem:
Score : 400 points

Problem Statement

Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left.
Each square is painted black or white.

The grid is said to be good if and only if the following condition is satisfied:

From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.

Note that (1, 1) and (H, W) must be white if the grid is good.

Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.

Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.

Constraints

2 \leq H, W \leq 100

Input

Input is given from Standard Input in the following format:

H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}

Here s_{rc} represents the color of (r, c) - # stands for black, and . stands for white.

Output

Print the minimum number of operations needed.

Sample Input 1

3 3
.##
.#.
##.

Sample Output 1

1

Do the operation with (r_0, c_0, r_1, c_1) = (2, 2, 2, 2) to change just the color of (2, 2), and we are done.

Sample Input 2

2 2
#.
.#

Sample Output 2

2

Sample Input 3

4 4
..##
#...
###.
###.

Sample Output 3

0

No operation may be needed.

Sample Input 4

5 5
.#.#.
#.#.#
.#.#.
#.#.#
.#.#.

Sample Output 4

4

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

import java.awt.Point;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Deque;
import java.util.NoSuchElementException;

public class Main implements Runnable {
    public static void main(String[] args) {
        new Thread(null,new Main(), "" ,Runtime.getRuntime().maxMemory()).start();
    }

    // メイン処理
    public void run() {
        FastScanner sc = new FastScanner();
        int h = sc.nextInt();
        int w = sc.nextInt();
        char[][] map = new char[h][w];
        for(int i = 0;i < h;i++) {
            char[] line = sc.next().toCharArray();
            for(int j = 0;j < w;j++) map[i][j] = line[j];
        }

        int[][] costMap = new int[h][w];
        for(int i = 0;i < h;i++) Arrays.fill(costMap[i], 10000);
        if(map[0][0] == '.') costMap[0][0] = 0;
        else costMap[0][0] = 1;
        int[] dx = {0, 1};
        int[] dy = {1, 0};
        boolean[][] footprints = new boolean[h][w];
        footprints[0][0] = true;
        Point start = new Point(0, 0);
        Deque<Point> que = new ArrayDeque<>();
        que.offer(start);
        while(!que.isEmpty()) {
            Point p = que.poll();
            for(int i = 0;i < 2;i++) {
                int nextX = p.x + dx[i];
                int nextY = p.y + dy[i];
                if(nextX >= w || nextY >= h) continue;

                char here = map[p.y][p.x];
                char nextField = map[nextY][nextX];
                int nextFieldCost = costMap[nextY][nextX];
                int hereCost = costMap[p.y][p.x];
                int candidateNextFieldCost;
                if(here == '.' && nextField == '.') candidateNextFieldCost = hereCost;
                else if(here == '.' && nextField == '#') candidateNextFieldCost = hereCost + 1;
                else if(here == '#' && nextField == '#') candidateNextFieldCost = hereCost;
                else candidateNextFieldCost = hereCost;
                if(candidateNextFieldCost < nextFieldCost) costMap[nextY][nextX] = candidateNextFieldCost;
                if(!footprints[nextY][nextX]) {
                    Point nextP = new Point(nextX, nextY);
                    if(nextField == '.') que.addFirst(nextP);
                    else que.addLast(nextP);
                    footprints[nextY][nextX] = true;
                }
            }
        }

        PrintWriter out =  new PrintWriter(System.out);
        out.println(costMap[h - 1][w - 1]);
        out.flush();
    }

    // 以下、ユーティリティ
    long nCm(long n, long m) {
        if(n < m) return 0l;
        long c = 1l;
        m = (n - m < m ? n - m : m);
        for(long ns = n - m + 1, ms = 1;ms <= m;ns ++, ms++) {
            c *= ns;
            c /= ms;
        }
        return c;
    }

    long modnCm(int n, int m, int mod) {
        if(n < m) return 0l;
        long upper = 1l;
        long downer = 1l;
        m = (n - m < m ? n - m : m);
        for(long ns = n - m + 1, ms = 1;ms <= m;ns ++, ms++) {
            upper = upper * ns % mod;
            downer = downer * ms % mod;
        }
        return upper * modInv(downer, mod) % mod;
    }

    // mod m での a の逆元 を計算する
    long modInv(long a, long m) {
        long b = m, u = 1, v = 0;
        while(b > 0) {
            long t = a / b;
            a -= t * b;
            long tmp = a;
            a = b;
            b = tmp;
            u -= t * v;
            tmp = u;
            u = v;
            v = tmp;
        }
        u %= m;
        if (u < 0) u += m;
        return u;
    }

    long modPow(long a, long n, long mod) {
        long res = 1l;
        while(n > 0l) {
            if(n % 2l == 1l) res = res * a % mod;
            a = a * a % mod;
            n /= 2l;
        }
        return res;
    }

}

class UnionFind {
    private int[] parent;

    UnionFind(int n) {
        parent = new int[n];
        for(int i = 0;i < n;i++) parent[i] = i;
    }

    public int rootOf(int x) {
        if(parent[x] == x) return x;
        else return parent[x] = rootOf(parent[x]);
    }

    public boolean same(int x, int y) {
        return rootOf(x) == rootOf(y);
    }

    public void unite(int x, int y) {
        x = rootOf(x);
        y = rootOf(y);
        if(x == y) return;
        parent[x] = y;
        return;
    }
}

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    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;}
    public boolean hasNext() { while(hasNextByte() && !isPrintableChar(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 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 double nextDouble() { return Double.parseDouble(next());}
}