bit 1015 Edit Distance

Edit Distance

时间限制: 1秒  内存限制: 64M

Problem Description

Given a string, an edit script is a set of instructions to turn itinto another string. There are
three kinds of operations in an edit script:

Add: Output one character. This instruction does not consume any characters
from the source string.

Delete: Delete one character. That is, consume one character from the source string and output nothing.

Modify: Modify one character. That is, consume one character from the source string and output a character.

Now, We define that the edit distance is the minimal number of operations to change string A into string B.
Given two strings, find the edit distance from string A to string B.

Input

There are several test cases in the input.

The first line of each test case is two integers n and m (1 <= n,m <= 1000), indicate the length of the two strings. Then two lines are string A and string B.

The input is ended by EOF.

Output

For each test cases, please output the edit distance from string A to string B.

Sample Input

1 2

a

aa

2 2

aa

ab

Sample Output

1

1

原来以为是求最长子序列一下wa了，后来发现不是，状态转移方程是：

当a[i-1]等于b[j-1]时，状态转移为res[i][j]=res[i-1][j-1]否则,res[i][j]=MIN(res[i-1][j-1]+1,res[i][j-1]+1,res[i-1][j]+1)。

#include <stdio.h>#include<string.h>#define min_l(a,b,c) a<b?(a<c?a:c):(b<c?b:c)int  a_len,b_len;int dis[1010][1010];void max_dis(char *a,char *b){//memset(dis,0,sizeof(int)*1010*1010);for(int i = 1; i <=a_len; ++i){for(int j = 1;j <=b_len;++j){if(!(a[i - 1] - b[j - 1])){dis[i][j] = dis[i-1][j-1];}else{dis[i][j] = min_l(dis[i - 1][j] + 1,dis[i][j-1] + 1,dis[i-1][j-1] + 1);}}}return ;}int main(int argc, char *argv[]){ char a[1100],b[1100];for(int i = 0; i < 1010;++i){dis[0][i] = i; //a 的长度为0 b的长度为i时的最大编辑距离 dis[i][0] = i; //b 的长度为0 a的长度为i时的最大编辑距离 }while(~scanf("%d%d",&a_len,&b_len)){scanf("%s%s",a,b);max_dis(a,b);printf("%d\n",dis[a_len][b_len]); }return 0;}