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Re:数据太弱,可以水过 0MS(贴有点问题但AC的代码)In Reply To:数据太弱,可以水过 0MS(贴有点问题但AC的代码) Posted by:szy532164656 at 2012-11-08 22:32:58 > #include<iostream> > #include<string> > using namespace std; > > int main() > { > int number; > string input; > while (true) > { > cin>>number; > if ( number == 0 ) > break; > int mark[26] = {0}; > mark[0] = 1; > bool relation[26][26] = {false}; > int maxMark = 1; > for ( int count = 0 ; count < number ; count ++ ) > { > cin>>input; > int lenght = input.length(); > int selfNum = input.at(0) - 'A'; > if ( lenght >= 3 ) > { > for ( int i = 2 ; i < lenght ; i ++ ) > { > int num = input.at(i) - 'A'; > relation[selfNum][num] = true; > } > } > } > for ( int i = 1 ; i < number ; i ++ ) > { > bool color[27] = {false}; > for ( int j = 0 ; j < number ; j ++ ) > { > > if ( relation[i][j] && i != j ) > { > color[mark[j]] = true; > } > } > for ( int j = 1 ; j <= number ; j ++ ) > { > if ( !color[j] ) > { > mark[i] = j; > break; > } > } > if ( mark[i] > maxMark ) > { > maxMark = mark[i]; > if ( maxMark >= 4 ) > { > maxMark = 4; > break; > } > } > } > if ( maxMark == 1 ) > cout<<maxMark<<" channel needed."<<endl; > else > cout<<maxMark<<" channels needed."<<endl; > } > } > > 6 > A:BEF > B:AC > C:BD > D:CEF > E:ADF > F:ADE > 结果为3 A(1)B(2)C(3)D(1)E(2)F(3) > 这组数据按以上算法得出结果4. Followed by: Post your reply here: |
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