字串距离

字串距离

题解

#include < fstream >
using   namespace  std;

int  n1,n2,k;
string  s1,s2;
int  f[ 2001 ][ 2001 ];
int  main()
{
    ifstream cin( " blast.in " );
    ofstream cout( " blast.out " );
    cin >> s1 >> s2 >> k;
    f[ 0 ][ 0 ] = 0 ;
    n1 = s1.size();
    n2 = s2.size();
     for ( int  i = 1 ;i <= n1; ++ i)
        f[i][ 0 ] = f[i - 1 ][ 0 ] + k;
     for ( int  i = 1 ;i <= n2; ++ i)
        f[ 0 ][i] = f[ 0 ][i - 1 ] + k;
     for ( int  i = 1 ;i <= n1; ++ i)
         for ( int  j = 1 ;j <= n2; ++ j){
             if (f[i - 1 ][j] + k > f[i][j - 1 ] + k)
                f[i][j] = f[i][j - 1 ] + k;
             else  f[i][j] = f[i - 1 ][j] + k;
             if (f[i][j] > abs(s1[i - 1 ] - s2[j - 1 ]) + f[i - 1 ][j - 1 ])
                f[i][j] = abs(s1[i - 1 ] - s2[j - 1 ]) + f[i - 1 ][j - 1 ];
        }
    cout << f[n1][n2] << endl;
     return   0 ;
}