POJ3356--AGTC（edit distance）动态规划

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 <cstdio>
#include <cstring>
using namespace std;
#define maxn 1080
char str1[maxn],str2[maxn];
int dp[maxn][maxn];
inline int min(int a,int b,int c)
{
	if(a>b) a = b;
	if(a>c) a = c;
	return a;
}
int main()
{
	int len1,len2;
	while(scanf("%d %s",&len1,str1+1)!=EOF)
	{
		scanf("%d %s",&len2,str2+1);
		dp[0][0] = 0;
		int maxlen = max(len1,len2);
		for(int i = 1;i <= maxlen;i++)
			dp[0][i] = dp[i][0] = i;
		for(int i = 1;i <= len1;i++)
			for(int j = 1;j <= len2;j++)
			{
				dp[i][j] = min(dp[i-1][j]+1,dp[i][j-1]+1,dp[i-1][j-1]+(str1[i]!=str2[j]));
			}
		printf("%d\n",dp[len1][len2]);
	}
	return 0;
}