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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.
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