hdu 2412 Party at Hali-Bula（树形DP）

Party at Hali-Bula

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1860    Accepted Submission(s): 649

Problem 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 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
  

 

Source

2006 Asia Regional Tehran

 

1520的加强版，比1520多增加了判断是否唯一，刚开始我竟然傻傻的判断是否有dp[i][0]==dp[i][1],其实儿子去与不去都一样，但是父亲去了，儿子就不能去了，就成了唯一的啦，再用一个dps[maxn][2]标记一下，是否值唯一，

1，如果儿子去不去都一样，父亲也不去，就有多种方案,；

2，如果儿子不去有多种方案，父亲去也有多种方案；

3，如果儿子不去有多种方案，并且，不去的人数大于去的，父亲不去也有多种方案,；

4，如果儿子去有多种方案，并且，取得人数大于不去的，父亲不去也有多种方案；、

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;
const int maxn=200+100;
int t;
int dp[maxn][2];
int dps[maxn][2];
vector<int>son[maxn];
map<string,int> mp;
void dfs(int u)
{
    dp[u][1]=1;
    for(int i=0;i<son[u].size();i++)
    {
        int v=son[u][i];
        dfs(v);
        dp[u][0]+=max(dp[v][0],dp[v][1]);
        dp[u][1]+=dp[v][0];
        if(dp[v][0]>dp[v][1]&&dps[v][0]==0)//如果儿子不去有多种方案，并且，不去的人数大于去的，父亲不去也有多种方案,
        dps[u][0]=0;
        if(dp[v][0]<dp[v][1]&&dps[v][1]==0)//如果儿子去有多种方案，并且，取得人数大于不去的，父亲不去也有多种方案
        dps[u][0]=0;
        if(dp[v][0]==dp[v][1])//如果儿子去不去都一样，父亲也不去，就有多种方案,
        dps[u][0]=0;
        if(dps[v][0]==0)//如果儿子不去有多种方案，父亲去也有多种方案
        dps[u][1]=0;
    }
    return;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==0)
        break;
        for(int i=0;i<=n;i++)
        son[i].clear();
        mp.clear();
        memset(dp,0,sizeof(dp));
        memset(dps,1,sizeof(dps));
        t=1;
        string s,s1;
        int x,y;
        cin>>s;
        mp[s]=t++;
        for(int i=1;i<n;i++)
        {
           cin>>s>>s1;
           if(!mp[s])
           {
               mp[s]=t++;
           }
           if(!mp[s1])
           {
               mp[s1]=t++;
           }
           x=mp[s];
           y=mp[s1];
           son[y].push_back(x);
        }
        dfs(1);
        if(dp[1][0]>dp[1][1]&&dps[1][0])
        {
            printf("%d Yes\n",dp[1][0]);
        }
        else if(dp[1][0]<dp[1][1]&&dps[1][1])
        {
            printf("%d Yes\n",dp[1][1]);
        }
        else
        {
            printf("%d No\n",max(dp[1][0],dp[1][1]));
        }
    }
    return 0;
}