sdutoj1225--Editing Distance (DP: string conversion)

Edit Distance

Nid=24#time "style=" padding-bottom:0px; margin:0px; padding-left:0px; padding-right:0px; Color:rgb (83,113,197); Text-decoration:none; padding-top:0px ">time limit:1000ms Memory limit:65536k have questions? Dot here ^_^

Description of the topic if the basic operation of a string is only: Delete a character, insert a character, and change one of the characters to one character. Three operations.

We have carried out a random operation of the above three operations called the basic operation of the one-step character.
Here we define the editing distance of two strings: for two strings A and B. Through the above basic operation. We can change A to B or B to a, so the minimum number of basic character steps required for string A into string B is called the editing distance of string A and string B.

For example: A= "ABC", b= "CBCD", the editing distance of A and B is 2.
Your task is to compile a high-speed program to calculate the editing distance of randomly two strings. The input input includes multiple sets of test data.

A row of test data for each group, string A and string B.

The length of the string is not greater than 1024. And they are all letters.

Output editing distance. Demo sample Input

ABC CBCD

Demo sample Output

2

Hint Source ZjgsuFrom a string to B string, three operations, written as two-dimensional DP, Dp[i][j] represents a string matching to the i,b string matching to J total number of steps required.

suppose s1[i] = = S2[j] So directly from the dp[i-1][j-1] to get. Or dp[i-1][j]+1 represents the addition of a character, dp[i][j-1]+1 represents the deletion of one character from theassumptions are not the same. Then, as with the above, only the change for dp[i-1][j-1]+1 represents a character change. #include <cstdio>
#include <cstring>
#include <algorithm>
using namespace Std;
Char s1[1200], s2[1200];
int dp[1200][1200];
int main ()
{
int I, J, L1, L2;
while (scanf ("%s%s", S1, S2)!=eof)
{
Memset (Dp,0,sizeof (DP));
L1 = strlen (S1);
L2 = strlen (s2);
for (i = 0; I <= L1; i++)
Dp[i][0] = i;
for (j = 0; J <= L2; j + +)
DP[0][J] = j;
for (i = 1; I <= L1; i++)
{
for (j = 1; J <= L2; j + +)
{
if (s1[i-1] = = S2[j-1])
Dp[i][j] = min (dp[i-1][j-1],min (dp[i-1][j]+1,dp[i][j-1]+1));
Else
Dp[i][j] = min (dp[i-1][j-1]+1,min (dp[i-1][j]+1,dp[i][j-1]+1));
}
}
printf ("%d\n", Dp[l1][l2]);
}
return 0;
}

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sdutoj1225--Editing Distance (DP: string conversion)