ACM 编辑距离

Description

Let x and y be two strings over some finite alphabet A. We would like to transform x into y allowing only operations given below:

• Deletion: a letter in x is missing in y at a corresponding position.
• Insertion: a letter in y is missing in x at a corresponding position.
• Change: letters at corresponding positions are distinct

Certainly, we would like to minimize the number of all possible operations.

Illustration

A G T A A G T * A G G C 
| | |       |   |   | |

A G T * C * T G A C G C

Deletion: * in the bottom line 
Insertion: * in the top line 
Change: when the letters at the top and bottom are distinct

This tells us that to transform x = AGTCTGACGC into y = AGTAAGTAGGC we would be required to perform 5 operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should do it like

A  G  T  A  A  G  T  A  G  G  C 
|  |  |        |     |     |  |

A  G  T  C  T  G  *  A  C  G  C

and 4 moves would be required (3 changes and 1 deletion).

In this problem we would always consider strings x and y to be fixed, such that the number of letters in x is m and the number of letters in y is n where n ≥m.

Assign 1 as the cost of an operation performed. Otherwise, assign 0 if there is no operation performed.

Write a program that would minimize the number of possible operations to transform any string x into a string y.

Input

The input consists of the strings x and y prefixed by their respective lengths, which are within 1000.

Output

An integer representing the minimum number of possible operations to transform any string x into a string y.

Sample Input

10 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

4

解题思路：

题目大意是将两个字符数组通过删除字符、插入字符、改变字符。来使这两个字符串变成相等的字符串。问我们最少需要经过多少步能使这两个字符串变成相等的字符串。

优子结构：dp[i][j]表示从a[i]到b[j]完全匹配的最小操作数

状态转移方程：

          1.dp[i][0]=i,dp[0][i]=i     //这是初始化步骤，这符合规律，因为这种情况下只能执行删除操作，

                 2.dp[i][j]=dp[i-1][j-1] (a[i]=b[j]) //相等无需变化，因此操作数也不增加

          3.dp[i][j]=min{dp[i-1][j]+1,dp[i][j-1]+1,dp[i-1][j-1]+1} (a[i]!=b[j])  //不相等还要考虑替换，插入          操作
程序代码：

#include <iostream>
using namespace std;
char a[1005],b[1005];
int dp[1005][1005];
int n1,n2;
int min(int x,int y)
{
	return x<y?x:y;
}
int maxlen(char *str1,char *str2)
{
	int i,j;
    //n1=strlen(str1),n2=strlen(str2);
	int len=n1>n2?n1:n2;
    for(i=0;i<=len;i++)
        dp[i][0]=dp[0][i]=i;
    for(i=1;i<=n1;i++)
	{
        for(j=1;j<=n2;j++)   //以下从str1[0]和str2[0]开始比较
		{
            if(a[i]==b[j])
				dp[i][j]=dp[i-1][j-1];
			else
				dp[i][j]=min( min(dp[i][j-1]+1,dp[i-1][j]+1),dp[i-1][j-1]+1);
		}
	}
    return dp[n1][n2];
}
int main()
{
    while(~scanf("%d%s",&n1,a+1))
     {
         scanf("%d%s",&n2,b+1);
         cout<<maxlen(a,b)<<endl;
     }
    return 0;
}