DP - 字符混编字符混编 Problem's Link

字符混编 

Problem's Link

 ----------------------------------------------------------------------------

Mean: 

略

analyse:

略

Time complexity: O(N)

 

view code

#include "bits/stdc++.h"
using namespace std;

class Mixture {
public :
    bool chkMixture( string A , int n , string B , int m , string C , int v) {
        if ( v != n + m)
            return false;
        bool dp [n + 1 ][ m + 1 ];
        memset( dp , 0 , sizeof dp);
        for ( int i = 1; i < n; ++ i)
            dp [ i ][ 0 ] = ( A [ i - 1 ] == C [ i - 1 ] ? 1 : 0);
        for ( int i = 1; i < m; ++ i)
            dp [ 0 ][ i ] = (B [ i - 1 ] == C [ i - 1 ] ? 1 : 0);
        for ( int i = 1; i <=n; ++ i) {
            for ( int j = 1; j <= m; ++ j) {
                dp [ i ][ j ] = ( A [ i - 1 ] == C [ i + j - 1 ] && dp [ i - 1 ][ j ]) || (B [ j - 1 ] == C [ i + j - 1 ] && dp [ i ][ j - 1 ]);
            }
        }
        cout << "------------------------------------------------------------" << endl;
        for ( int i = 0; i <= n; ++ i)
        {
            for ( int j = 0; j <= m; ++ j)
            {
                cout << dp [ i ][ j ] << " ";
            }
            cout << endl;
        }
        cout << "------------------------------------------------------------" << endl;
        return dp [n ][ m ];
    }
};

int main( int argc , char const * argv []) {
    string s1 , s2 , s3;
    while ( cin >> s1 >> s2 >> s3) {
        Mixture solution;
        if ( solution . chkMixture( s1 , s1 . length (), s2 , s2 . length (), s3 , s3 . length()))
            cout << "Yes." << endl;
        else
            cout << "No." << endl;
    }
    return 0;
}