Dynamic programming--c editing shortest distance

C-Edit Distance time limit: Ms . Memory Limit: 65536KB 64-bit input/output format: %i64d &%I 64u SubmitStatus

Describe

Let x and y are both strings over some finite alphabet A. We would like to transform x into y allowing only operations given below:

• deletion: a letter with x is missing in y at a corresponding position.
• insertion: a letter with y is missing in x at a corresponding position.
• Change : letters at corresponding positions is distinct

Certainly, we would like to minimize the number of all possible operations.

Illustration

| | |       |   |   | |
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 is distinct

This tells us, transform x = AGTCTGACGC into y = AGTAAGTAGGC We would being required to perform 5 Operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should does it like

A  g  t  a  a  g  t  a  g  g  
|  |  |        |     |     |  |
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 all consider strings x and y to is fixed, such that number of letters in x is m and the number of letters in yis 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 This 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 is within 100 0.

Output

An integer representing the minimum number of possible operations to transform any string x into a string Y.

Sample Input

Ten AGTCTGACGC11 AGTAAGTAGGC

Sample Output

4
The main topic: Give two strings x, Y, to find out the minimum number of operations from X-->y, can only delete, add, modify a character.
Problem Solving Ideas:

is also a classic problem in DP

D[I][J] represents the first string to the I position, and the second string to the J position of the shortest editing distance

D[I][J]

If A[I]==B[J]

D[i][j]=min (D[i-1][j-1],d[i-1][j]+1,d[i][j-1]);

Otherwise d[i][j]=min (d[i-1][j-1]+1,d[i-1][j]+1,d[i][j-1]);

Program code:

1#include <cstdio>2#include <iostream>3 using namespacestd;4 Const intm=1100;5 CharA[m],b[m];6 intD[m][m];7 intMain ()8 {9     intN,i,j,l,r;Ten     while(SCANF ("%d%s", &l,a+1)!=EOF) One    { Ascanf"%d%s", &r,b+1); -    intlen=Max (l,r); -      for(i=0; i<=len;i++) the        { -c[n][0]=i; -d[0][i]=i; -        } +      for(i=1; i<=l;i++) -          for(j=1; j<=r;j++) +         { AD[i][j]=min (d[i-1][j]+1, d[i][j-1]+1); at             if(a[i]==B[j]) -d[i][j]=d[i-1][j-1]; -             Else -D[i][j]=min (d[i][j],d[i-1][j-1]+1); -         } -printf"%d\n", D[l][r]); in    } -     return 0; to}

View Code

Dynamic planning--c editing minimum distance