AGTC

Let x and y be two strings over some finite alphabet A. We would like to transformx 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 inx is m and the number of letters in y is n wheren ≥ 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 stringx 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 stringx into a string y.

Sample Input

10 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

4

题意：将字符串s1通过删除，添加和变换变成s2的最小操作数

题解：dp[i][j]的意义为y取前i个字母和x取前j个字母的最少操作次数

#include<iostream>
#include<algorithm>
#include<string.h>
char s1[1005];
char s2[1005];
int dp[1005][1005];
using namespace std;
int main(){
	int n;
	int m;
	while(cin>>n>>s1>>m>>s2){
		memset(dp,0,sizeof(dp));
		//s1经过i次删除与s2的前0个字符 
		for(int i=1;i<=n;i++)
			dp[i][0]=i;
		//s1经过j次添加与s2的前j个字符 
		for(int j=1;j<=m;j++)
			dp[0][j]=j;
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++){ 
				if(s1[i-1]==s2[j-1])
					//若字符相等，则无需操作
					dp[i][j]=dp[i-1][j-1];
				else
					//x删除一个字符，x增加一个字符（相当于j删除一个字符） 
					dp[i][j]=min(dp[i-1][j]+1 ,dp[i][j-1]+1);
				dp[i][j]=min(dp[i][j],dp[i-1][j-1]+1); 
			} 
		cout<<dp[n][m]<<endl;
	}
	return 0;
}