动态规划中，在给定两个字符串a和b的情况下，在允许进行删除、插入或替换操作的基础上求解将一个字符串转换为另一个字符串所需的最小操作次数

    #include <iostream>
    #include <string>
    using namespace std;
    
    const int Max = 1000;
    char a[Max],b[Max];
    int f[Max][Max];
    int m , n;
    
    int main (void) {
    	int n , m ;
    	cin>>n>>m>>a+1>>b+1;
    	for (int i = 0; i <=m; i++) {//colmn 0 and row 0 means the NULL transforms into the target string.
    		f[i][0]=i;
    	}
    	for (int j = 0; j <=n; j++) {
    		f[0][j]=j;
    	} 
    	
    	/*the equation: f(x,y)= {
    		colmn 0 and row 0 = i or j , 					i=0 or j = 0;
    		min( f(i-1,j)+1, f(i,j-1)+1, f(i-1,j-1)+