DP Editing Distance problem

Test instructions

The editing distance problem is to give two strings;
Such as:
ABCdef
Abbefc
The second string becomes the first string by adding, deleting, and changing three ways;
It is obvious that this example is a 3-time change;
The minimum number of changes to a given string after the topic is asked;

Understand:

A look at this problem is actually covered;
Of course there is a way to do, that is, wide search;
But the wide search in time is not solve the problem;
So we have to seek another way out;
The real concrete problem-solving idea is as follows:
First of all:
We add some underscores to two strings;
The
Abcdef_
Ab_befc
Can be seen:
By adding or deleting changes can obviously transform the past;
So what is the case, what is changed, and what is the deletion of it?
We now have a minimum value f (i, j) that needs to be changed;
represents the smallest change that exists between the I of the first string and the J character of the second string;
We know if str1[i] = = Str2[j] words;
Then: F (i, j) = f (i-1, j-1);
if not equal;
Then: F (i, j) = f (i-1, j-1) + 1;
This is the change;
And the first I character is the same as the former j-1 characters, then it is necessary to delete;
Such as:
Abab
Ababc
Then: F (i, j) = f (i, j-1) + 1;
And the first i-1 characters in the first J characters are the same, then add;
Ababa
Abab
Then: F (i, j) = f (i-1, J) + 1;
The last two are difficult to understand, but indispensable;
Thus, we can find a minimum value by the above three formulas;
The answer is the minimum value;

The code is as follows:

#include <cstdio> #include <cstring> #include <cmath> #include <ctime> #include <iostream > #include <algorithm> #include <vector> #include <string> #include <map> #include <set

> #include <queue> #include <stack> using namespace std;
typedef long Long LL;

typedef pair<int, int> PII;
Const double Min_inf = 1e-7;

const int max_inf = (1e9) + 7;
    #define X First #define Y second int main () {string str1, str2;
    CIN >> str1 >> str2;

    int len1 = Str1.length (), len2 = Str2.length ();
    Vector<vector<int> > DP (LEN1 + 2, vector<int> (len2 + 2, 0));
    for (int i = 1; I <= len1; ++i) {//initial value dp[i][0] = i;
    } for (int i = 1; I <= len2; ++i) {//initial value dp[0][i] = i; } for (int i = 1, i <= len1; ++i) {for (int j = 1; j <= Len2; ++j) {if (str1[i-1] = = St
   R2[j-1]) {dp[i][j] = dp[i-1][j-1];         } else {Dp[i][j] = min (dp[i-1][j-1], min (Dp[i-1][j], dp[i][j-1])) + 1;

    }}} cout << dp[len1][len2] << Endl;
return 0;
 }