Schedule zoj 3348

Schedule

Time Limit: 1 Second Memory Limit: 32768 KB

DD enters the top-level pingpong match! As the game going on, he wonders if there is possibility for him to get the champion(excluding joint champions). Now provide you the results of previous matches, and the details of rest matches. You have to judge it.

Input

There are several test cases. You should process to the end of input.

For each test case. the first line contain two integers n (1 <= n <= 50) and m(0 <= m <= 1000), n indicates the number of players. m indicates the matches have ended.

Then for the following m lines, each line has the format: "Player1 Player2 Result". Player1 and Player2 are the names of the two players, and Result could only be "win" or "lose".

In the next line there is another integer p (0 <= p <= 5000), indicates the matches that will start later.

Then for the following p lines, each line has the format: "Player1 Player2", which means Player1 and Player2 will start a match.

We ensure there is a player named "DD". The length of each player name is no longer than 10.

Output

You should only output "Yes" or "No" for each case, which means if DD has the possibility to be the champion.

Sample Input

3 2
DD winsty win
DD owen lose
2
winsty owen
DD owen
3 3
DD winsty win
DD owen lose
winsty owen lose
2
owen winsty
owen DD

Sample Output

Yes
No

很经典的构图，最大流！

先统计每个人已经赢的次数和假设比赛他赢的次数，记c[i]

S连到每个赛手，流量为C[i]

……

int Getid(string name)
{
	int i,j;
	for(i=1;i<=alr;i++)
	{
		if(name==person[i])
			return i;
	}
	//cannot find such name
	person[++alr]=name;
	return alr;
}

void init()
{
	memset(p, -1, sizeof(p));
	memset(champ,0,sizeof(champ));
	memset(c,0,sizeof(c));
	ans = indexx = 0;
	int i,j,p1,p2,dd=0;
	string na1,na2,status;
	alr=0;
	for(i=0;i<temp;i++)
	{
		cin>>na1>>na2>>status;
		if(status[0]=='w')
		{
			if(na1=="DD")
				dd++;
			else
			{
				p1=Getid(na1);
				c[p1]++;
				ans++;
			}
		}
		else
		{
			if(na2=="DD")
				dd++;
			else
			{
				p2=Getid(na2);
				c[p2]++;
				ans++;
			}
		}
	}
	cin>>temp;
	for(i=0;i<temp;i++)
	{
		cin>>na1>>na2;
		//suppose first people win
		if(na1=="DD"||na2=="DD")
		{dd++;continue;}
		p1=Getid(na1);
		p2=Getid(na2);
		c[p1]++;
		champ[p1][p2]++;
		ans++;
	}
	for(i=1;i<=n;i++)
	{
		add(0,i,c[i]);
		add(i,n+1,dd-1);
		for(j=1;j<=n;j++)
			add(i,j,champ[i][j]);
	}
}