#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.

• Re:数据太弱，可以水过 0MS（贴有点问题但AC的代码） (1437) xnxylxh 2012-11-22 01:37:28