Problem:
Edit Distance (Levenshtein Distance)

Find the edit distance between given two words s1 and s2.

The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations:

insertion: Insert a character at a particular position.

deletion: Delete a character at a particular position.

substitution: Change the character at a particular position to a different character

Input

s1
s2

Two words s1 and s2 are given in the first line and the second line respectively. The words will consist of lower case characters.

Output

Print the edit distance in a line.

Constraints

1 ≤ length of s1 ≤ 1000

1 ≤ length of s2 ≤ 1000

Sample Input 1

acac
acm

Sample Output 1

2

Sample Input 2

icpc
icpc

Sample Output 2

0

Solution in C# (judge verdict: Accepted)

using System;
using System.Net.NetworkInformation;

public class Program
{
    public void Solve()
    {
        var sc = new Scanner();
        string s1 = sc.Next();
        string s2 = sc.Next();

        var dp = new int[s1.Length + 1, s2.Length + 1];
        for (int i = 1; i <= s1.Length; i++)
        {
            dp[i, 0] = i;
        }

        for (int j = 1; j <= s2.Length; j++)
        {
            dp[0, j] = j;
        }

        for (int i = 0; i < s1.Length; i++)
        {
            for (int j = 0; j < s2.Length; j++)
            {
                int a = dp[i + 1, j] + 1;
                int b = dp[i, j + 1] + 1;
                int c = dp[i, j] + (s1[i] == s2[j] ? 0 : 1);
                dp[i + 1, j + 1] = Math.Min(a, Math.Min(b, c));
            }
        }

        Console.WriteLine(dp[s1.Length, s2.Length]);
    }

    public static void Main(string[] args)
    {
        new Program().Solve();
    }
}

class Scanner
{
    public Scanner()
    {
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = Console.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = Console.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}