"Codevs 2598" editing distance problem

2598 Editing distance problems
Time limit: 1 s
Space limit: 128000 KB
Title Level: Diamonds Diamond
Exercises
Title Description Description
Set A and B are 2 strings. To convert string A to string B with a minimum of character manipulation. The character operations described here include:

(1) Delete a character;

(2) inserting a character;

(3) Change one character to another character.

The minimum character operand used to transform string A into string B is called the editing distance of the string A through B, which is recorded as D (A, A, a). Try to write the program, on the given 2 strings A and B, calculate their editing distance d (A, a, a, a,).

Enter a description input Description
The input file edit.in has two lines, the first line is string A, and the second line is string B.

Outputs description Output Description
The output file edit.out only one line, which is the editing distance d (A, b).

Sample input to sample
Fxpimu

Xwrs

Sample output Sample Outputs
5

Data Size & Hint
The length of the 40% data string A and B is not more than 100;

The length of the 100% data string A, B is no more than 4000.

DP[I][J] The editing distance of a string before I and B string before J characters

Dp[i][j] = i; (j = = 0)
Dp[i][j] = j; (i = = 0)
Dp[i][j] = min (Dp[i-1][j] + 1,dp[i][j-1] + 1,dp[i-1][j-1] + 1); (A[i]! = B[i])
Dp[i][j] = dp[i-1][j-1] + 1; ( A[i] = = B[i])

 #include <iostream> #include <cstdio> #include <cstring> #include <
Cmath> #include <algorithm> using namespace std;
const int MAXN = 4002;
String A, B;
Char SA[MAXN],SB[MAXN];
int DP[MAXN][MAXN];
    int main () {memset (dp,0x3f,sizeof (DP));
    Cin >> A;
    CIN >> B;
    int n = a.length ();
    int m = B.length ();
    for (int i = 1; I <= n; i + +) sa[i] = a[i-1];
    for (int i = 1; I <= m; i + +) sb[i] = b[i-1];
    Dp[0][0] = 0;
    for (int i = 1; I <= n; i + +) dp[i][0] = i;
    for (int i = 1; I <= m; i + +) dp[0][i] = i;
                for (int i = 1; I <= n; i + +) {for (int j = 1; j <= M; j + +) {if (Sa[i]! = Sb[j])
            Dp[i][j] = min (dp[i-1][j-1] + 1,min (dp[i][j-1] + 1,dp[i-1][j] + 1));
        else dp[i][j] = dp[i-1][j-1];
    }} printf ("%d\n", Dp[n][m]);
return 0; }