Dear Contestant,

I'm going to have a party at my villa at Hali-Bula to celebrate my retirement from BCM. I wish I could invite all my co-workers, but imagine how an employee can enjoy a party when he finds his boss among the guests! So, I decide not to invite both an employee and his/her boss. The organizational hierarchy at BCM is such that nobody has more than one boss, and there is one and only one employee with no boss at all (the Big Boss)! Can I ask you to please write a program to determine the maximum number of guests so that no employee is invited when his/her boss is invited too? I've attached the list of employees and the organizational hierarchy of BCM.

Best,
--Brian Bennett

P.S. I would be very grateful if your program can indicate whether the list of people is uniquely determined if I choose to invite the maximum number of guests with that condition.

Input

The input consists of multiple test cases. Each test case is started with a line containing an integer n

 (1 ≤ n ≤ 200), the number of BCM employees. The next line contains the name of the Big Boss only. Each of the following n-1 lines contains the name of an employee together with the name of his/her boss. All names are strings of at least one and at most 100 letters and are separated by blanks. The last line of each test case contains a single 0.

Output

For each test case, write a single line containing a number indicating the maximum number of guests that can be invited according to the required condition, and a word Yes or No, depending on whether the list of guests is unique in that case.

Sample Input

6
Jason
Jack Jason
Joe Jack
Jill Jason
John Jack
Jim Jill
2
Ming
Cho Ming
0

Sample Output

4 Yes
1 No

解題報告：

非常明顯的一樹形dp問題，至於要判斷有沒有重複，就只需要加點判斷。首先如果根節點選和不選的值一樣，那麼肯定就是有沒有唯一解，其次如果中間某個結點的值，由他的子節點的值推出來，而他的子節點已經被標記過了（就是子節點的最優解是有多種方式出來的），因此這個點的選擇方式也不說唯一的，而根節點處的最優解又用到了這個結點的值，那麼就是沒有唯一解。mk陣列就是用來記錄中間結點的最優解是不是唯一的。

ac程式碼：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<set>
#include<iostream>
#include<map>
#include<stack>
#include<cmath>
#include<algorithm>
#define ll long long
#define mod 1000000007
#define eps 1e-8
using namespace std;
struct node{
	int u;int v;
	int nex;
}side[22000];
int head[2000];
int dp[2000][2];
int mk[2000][2];
int cnt;
void init()
{
	cnt=0;
	memset(head,-1,sizeof(head));
	memset(dp,0,sizeof(dp)); 
	memset(mk,0,sizeof(mk));
}
void add(int u,int v)
{
	side[cnt].u=u;
	side[cnt].v=v;
	side[cnt].nex=head[u];
	head[u]=cnt++;
}
void dfs(int rt)
{
	for(int i=head[rt];i!=-1;i=side[i].nex)
	{
		int ty=side[i].v;
		dfs(ty);
		if(mk[ty][0]==1)
		{
			mk[rt][1]=1;
		}
		dp[rt][1]+=dp[ty][0];
		
		if(dp[ty][0]==dp[ty][1])
		{
			dp[rt][0]+=dp[ty][0];
			mk[rt][0]=1;
		}
		else if(dp[ty][0]>dp[ty][1])
		{
			dp[rt][0]+=dp[ty][0];
			if(mk[ty][0]==1)
			mk[rt][0]=1;
		}
		else
		{
			dp[rt][0]+=dp[ty][1];
			if(mk[ty][1]==1)
			mk[rt][0]=1;
		}
	}
}
map<string,int> mp;
int main()
{
	int n;
	while(~scanf("%d",&n),n)
	{
		mp.clear();
		init();
		int id=1;
		string s1,s2;
		int u,num,v;
		for(int i=0;i<n+1;i++)
		dp[i][1]=1;
		cin>>s1;
		mp[s1]=1;
		id++;
		for(int i=0;i<n-1;i++)
		{
			cin>>s1>>s2;
			if(!mp[s1])
			{
				mp[s1]=id++;
			}
			if(!mp[s2])
			{
				mp[s2]=id++;
			}
			add(mp[s2],mp[s1]);
		}
		int ans;
		int ff=1;
		dfs(1);
		if(dp[1][0]==dp[1][1])
		{
			ans=dp[1][0];
			ff=0;	
		}
		else if(dp[1][0]>dp[1][1])
		{
			ans=dp[1][0];
			if(mk[1][0])
			ff=0;
		}
		else
		{
			ans=dp[1][1];
			if(mk[1][1])
			ff=0;
		}
		if(ff)
		{
			printf("%d Yes\n",ans);
		}
		else
		{
			printf("%d No\n",ans);
		}		
	} 
	return 0;
}