学步园

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.

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.

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.

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

4 Yes
1 No
题意：第一行给出一个数n，表示n个人形成一棵关系树，第二行给出一个人名表示的是根节点，接下来n-1行描述的是两人有直接的上下级关系。问找出参加party的人中任意两人之间没有直接的上下级关系的最多人数，并且判断组成的人数是否唯一。
#include<stdio.h>
#include<iostream>
#include<map>
#include<string>
using namespace std;
struct nnn
{
    int k;
    int son[205];
}node[205];
int vist[205],dp[205][2],only[205][2];//dp[p][i]，i=0表示节点p不参与，i=1表示参与
int max(int a,int b)
{
    return a>b?a:b;
}
void dfs(int p)
{
    dp[p][0]=0; dp[p][1]=1; vist[p]=1;
    only[p][0]=only[p][1]=1;//表示唯一
    for(int i=1;i<=node[p].k;i++)
    {
        int son=node[p].son[i];
        if(vist[son])continue;
        dfs(son);

        dp[p][1]+=dp[son][0];
        if(!only[son][0])
                only[p][1]=0;
        dp[p][0]+=max(dp[son][0],dp[son][1]);
        if(dp[son][0]>dp[son][1]&&only[son][0]==0)only[p][0]=0;
        if(dp[son][0]<dp[son][1]&&only[son][1]==0)only[p][0]=0;
        if(dp[son][0]==dp[son][1]) only[p][0]=0;
    }
}
int main()
{
    int n,m,a,b,k,maxsum;
    char str[2][150];
    map<string,int>name;
    while(scanf("%d",&n)>0&&n)
    {
        m=0; name.clear();
        for(int i=0;i<=n+2;i++)
        vist[i]=0,node[i].k=0;

        scanf("%s",str[0]);
        m++; name[str[0]]=m;
        for(int i=1;i<n;i++)
        {
            scanf("%s%s",str[0],str[1]);
            if(!name[str[0]])
            {
                m++; name[str[0]]=m;
            }
            if(!name[str[1]])
            {
                m++; name[str[1]]=m;
            }
            a=name[str[0]]; b=name[str[1]];
            node[a].k++; k=node[a].k; node[a].son[k]=b;
            node[b].k++; k=node[b].k; node[b].son[k]=a;
        }
        dfs(1);
        int onl=1;
        maxsum=max(dp[1][0],dp[1][1]);
        if(dp[1][0]>dp[1][1]&&only[1][0]==0)onl=0;
        if(dp[1][0]<dp[1][1]&&only[1][1]==0)onl=0;
        if(dp[1][0]==dp[1][1]) onl=0;

        printf("%d %s\n",maxsum,onl?"Yes":"No");
    }
}