AGTC(动态规划-最短编辑距离)

AGTC

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10742 Accepted: 4133

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

Source

Manila 2006

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define N 1010int dp[N][N];//dp[i][j]表示a的前i个和b的前j个相同后的最短距离char a[N],b[N];void match(char *s,char *s1,int lena,int lenb){    int i,j,inserts,deletes,replaces;    for(i=1;i<=lena;i++)    {        dp[i][0]=i;    }    for(j=1;j<=lenb;j++)    {        dp[0][j]=j;    }    for(i=1;i<=lena;i++)    {        for(j=1;j<=lenb;j++)        {            inserts=dp[i][j-1]+1;//插入            deletes=dp[i-1][j]+1;//删除            replaces;//替换            if(s[i-1]==s1[j-1])            {                replaces=dp[i-1][j-1];            }            else            {                replaces=dp[i-1][j-1]+1;            }            dp[i][j]=min(min(inserts,deletes),replaces);        }    }    cout<<dp[lena][lenb]<<endl;}int main(){    int la,lb;    while(cin>>la>>a)    {        cin>>lb>>b;        match(a,b,la,lb);    }    return 0;}