Party at Hali-Bula - POJ 3342 树形dp

Party at Hali-Bula

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5200		Accepted: 1850

Description

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 followingn-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

题意：给你N组人，第一组只有一个人，是最大的Boss，后面的组，每组的后一个人是前一个人的老板，并且能够形成一个树形。不能同时邀请有直接关系的老板和雇员，问你最多邀请多少人，并且如果只有一种邀请方式输出Yes。

思路：dp[i][0]表示第i个人不参加的最大人数,dp[i][1]表示第i个人参加的最大人数。way[i][0],way[i][1]分别表示其最大人数的情况下，邀请方式是不是唯一的。然后可以通过雇员来推出他的老板的状态。具体的转移方程见代码吧。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<map>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
vector<int> vc[210];
map <string ,int > match;
int dp[210][2];
bool way[210][2];
char s[210];
void dfs(int a)
{ int i,j,k;
  dp[a][0]=0;
  dp[a][1]=1;
  for(i=0;i<vc[a].size();i++)
  { k=vc[a][i];
    dfs(k);
    dp[a][0]+=max(dp[k][0],dp[k][1]);
    if(dp[k][0]>dp[k][1])
    { if(way[k][0])
       way[a][0]=1;
    }
    else if(dp[k][0]<dp[k][1])
    { if(way[k][1])
       way[a][0]=1;
    }
    else
     way[a][0]=1;
    dp[a][1]+=dp[k][0];
    if(way[k][0])
     way[a][1]=1;
  }
  //printf("way %d %d %d\n",a,way[a][0],way[a][1]);
}
int main()
{ int n,i,j,k,u,v,num=1,ans;
  while(~scanf("%d",&n) && n>0)
  { for(i=0;i<=n;i++)
     vc[i].clear();
    memset(dp,0,sizeof(dp));
    memset(way,0,sizeof(way));
    match.clear();
    scanf("%s",s);
    match[s]=1;
    num=1;
    for(i=2;i<=n;i++)
    { scanf("%s",s);
      if(match[s]==0)
       match[s]=++num;
      u=match[s];
      scanf("%s",s);
      if(match[s]==0)
       match[s]=++num;
      v=match[s];
      vc[v].push_back(u);
    }
    dfs(1);
    ans=max(dp[1][0],dp[1][1]);
    if(dp[1][0]>dp[1][1])
    { if(way[1][0])
       printf("%d No\n",ans);
      else
       printf("%d Yes\n",ans);
    }
    else if(dp[1][0]<dp[1][1])
    { if(way[1][1])
       printf("%d No\n",ans);
      else
       printf("%d Yes\n",ans);
    }
    else
     printf("%d No\n",ans);
  }
}