Party at Hali-Bula POJ - 3342（树形dp）-白红宇

Party at Hali-Bula POJ - 3342（树形dp）

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发布时间：2019-05-25

本文共 2423 字，大约阅读时间需要 8 分钟。

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

Jason

Jack Jason

Joe Jack

Jill Jason

John Jack

Jim Jill

Ming

Cho Ming

Sample Output

4 Yes

1 No

这个题目和那个没有上司的舞会很相似，如果不是需要判断有没有固定的解，直接复制过来就好了，但是要判断是不是解的个数相同。这就需要好好判断一下了。dfs跑完以后，就需要每个点去遍历一下。如果当前点上司不去大于上司去，就要遍历一下它的子节点看看是否相同。

代码如下：

#include
   
    #include
    
     #include
     
      #include
      #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;map
           
            mp;vector
            
             vec[205];int dp[205][2];int n;char a[205],b[205];int cnt;char str[205][105];void dfs(int top){ for(int i=0;i
             
              >a; mp[a]=cnt++; for(int i=1;i
              
               >a>>b; /*if(!mp[a]) mp[a]=cnt++; if(!mp[b]) mp[b]=cnt++; vec[mp[b]].push_back(mp[a]);*/ if(mp.find(a)==mp.end()) mp[a]=cnt++;//很重要 if(mp.find(b)==mp.end()) mp[b]=cnt++; vec[mp[b]].push_back(mp[a]); } dfs(0); int flag=1; for(int i=0;i
               
                dp[i][1]) { for(int j=0; j

努力加油a啊，(^o)/~

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