poj3356(LCS)

AGTC

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12190 Accepted: 4575

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 AGTCTGACGC11 AGTAAGTAGGC

Sample Output

4

//poj3356(LCS)//题目大意:给你两个DNA序列A、B,A串的长度小于等于B串.现在然你通过插入、删除、以及修改//使得A串变为B串,问你最少需要多少次操作？//解题思路：首先可以先求出两串的最长公共子序列.这些不需要任何操作.然后由于使A转化为B,B的长度大于等于A.//所以B串的长度减去其最长公共子序列的长度就是最少需要的操作数.(这一点我是看样例想到的,一开始以为kmp) #include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[1010][1010];char s[1010],t[1010];int main(){   int n,m,i,j,k=0;   while(scanf("%d",&n)!=EOF)   {   scanf("%s",s);   scanf("%d",&m);   scanf("%s",t);   memset(dp,0,sizeof(dp));     for(i=0;i<n;i++)     {       for(j=0;j<m;j++)   {     if(s[i]==t[j]) dp[i+1][j+1]=dp[i][j]+1; else dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);   }     }   k=m-dp[n][m];   printf("%d\n",k);   }   return 0;}