POJ 3356 AGTC LCS变形题



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

#include<iostream>
#include <cstring>
#include<algorithm>
#include <cstdio>
#include <map>
#include <string>
using namespace std;
const int maxn=1000+10;
int dp[maxn][maxn];
int min3(int a,int b,int c)
{
	a=min(a,b);
	a=min(a,c);
	return a;
}
int main()
{
	int len1,len2;
	while(scanf("%d",&len1)!=EOF)
	{
		char s1[maxn];
		char s2[maxn];
		scanf("%s",s1);
		scanf("%d",&len2);
		scanf("%s",s2);
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=len1;i++)
		{
			dp[i][0]=dp[i-1][0]+1;
		}
		for(int i=1;i<=len2;i++)
		{
			dp[0][i]=dp[0][i-1]+1;
		}
		dp[0][0]=0;
		for(int i=1;i<=len1;i++)
		{
			for(int j=1;j<=len2;j++)
			{
				dp[i][j]=min3(dp[i-1][j-1]+((s1[i-1]==s2[j-1])?0:1),dp[i][j-1]+1,dp[i-1][j]+1);
			}
		}
		printf("%d\n",dp[len1][len2]);
	}
	return 0;
}