POJ 3356 AGTC (minimum editing distance), pojagtc

POJ 3356 AGTC (minimum editing distance), pojagtc

POJ 3356 AGTC (minimum editing distance)

Http://poj.org/problem? Id = 3356

Question:

Two strings x and y are given. The length of x is n, the length of y is m, and m> = n. then, y can delete a letter, add a letter, and convert a letter to x. how many operations can be performed at least?

Analysis:

We set dp [I] [j] = x to indicate that the first I characters of the source string must be x-step.

Initialization: Dp [0] [I] = I and dp [I] [0] = I.

The former indicates that the source string is added with I characters, and the latter indicates that the source string is deleted with I characters.

Status Transfer: Dp [I] [j] If weOnly at the end of the source stringPerform the following three operations.

1. dp [I-1] [J-1] + (x [I] = y [j]? 0: 1). If the source string and the object string last character is the same, natural dp [I-1] [J-1] is to ask. If they last character is different, then you canReplaceThe last character of the source string is the character of the object string.

2. dp [I-1] [j] + 1. We canDeleteThe last character of the source string (step 1 delete operation), and then the source string into the object string (dp [I-1] [j] Step operation ).

3. dp [I] [J-1] + 1. We can convert the source string to the string of the first J-1 character (Dp [I] [J-1] Step operations), And then at the end of the source stringAddA y [j] becomes the object string (Step 1).

Final Demand: Minimum values of the preceding three formulas.

AC code:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1000+5;int n,m;int dp[maxn][maxn];char s1[maxn],s2[maxn];int main(){    while(scanf("%d%s",&n,s1)==2)    {        scanf("%d%s",&m,s2);        for(int i=0;i<=m;i++)            dp[0][i]=i;        for(int i=0;i<=n;i++)            dp[i][0]=i;        for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++)        {            dp[i][j]=dp[i-1][j-1]+(s1[i-1]==s2[j-1]?0:1);            dp[i][j]=min(dp[i][j], min(dp[i-1][j]+1,dp[i][j-1]+1));        }        printf("%d\n",dp[n][m]);    }    return 0;}

Please describe the problem