Party at Hali-Bula（UVA - 1220）

Party at Hali-Bula

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.

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这道题和树的最大独立集非常相似，但是加了一个条件，就是判断唯一性；

最大独立集就不说了，式子为：

dp[i][1]+=dp[j][0];
dp[i][0]+=max(dp[j][1],dp[j][0]); // j 为 i 的子节点

那么这个唯一性怎么判断呢？我们可以用一个标记数组f[][]来表示，f[i][1|0]=1表示这个方案唯一，f[i][1|0]=0表示这个方案不唯一；

当计算dp[i][1]时，只要一个子节点f[j][0]=0，则f[i][1]=0,也就是一个子节点不唯一，那么就不唯一；

计算dp[i][0]也是这样；

代码：

#include<bits/stdc++.h>
#define LL long long
#define pa pair<int,int>
#define lson k<<1
#define rson k<<1|1
#define inf 0x3f3f3f3f
//ios::sync_with_stdio(false);
using namespace std;
const int N=10100;
const int M=4001000;
const LL mod=998244353;
struct Node{
	int to,nex;
}edge[N<<1];
int head[N];
int cnt;
void add(int p,int q){
	edge[cnt].to=q;
	edge[cnt].nex=head[p];
	head[p]=cnt++;
}
bool vis[N];
int dp[N][2];
int f[N][2];
void dfs(int p){
	vis[p]=true;
	dp[p][1]=1;
	dp[p][0]=0;
	f[p][1]=1;
	f[p][0]=1;
	for(int i=head[p];~i;i=edge[i].nex){
		int q=edge[i].to;
		if(vis[q]) continue;
		dfs(q);
		dp[p][1]+=dp[q][0];
		if(f[q][0]==0) f[p][1]=0;
		if(dp[q][1]==dp[q][0]){
			f[p][0]=0;
			dp[p][0]+=dp[q][1];
		}
		if(dp[q][1]>dp[q][0]){
			if(f[q][1]==0) f[p][0]=0;
			dp[p][0]+=dp[q][1];
		}
		if(dp[q][1]<dp[q][0]){
			if(f[q][0]==0) f[p][0]=0;
			dp[p][0]+=dp[q][0];
		}
	}
	return;
} 
int main(){
	ios::sync_with_stdio(false);
	int n;
	while(1){
		memset(head,-1,sizeof(head));
		memset(vis,false,sizeof(vis));
		cnt=0;
		set<string>se;
		map<string,int>ma;
		cin>>n;
		if(n==0) break;
		string s1,s2;
		int ans=1;
		cin>>s1;
		se.insert(s1);
		ma[s1]=ans;
		for(int i=1;i<n;i++){
			cin>>s1>>s2;
			if(!se.count(s1)){
				se.insert(s1);
				ma[s1]=++ans;
			}
			if(!se.count(s2)){
				se.insert(s2);
				ma[s2]=++ans;
			}
			add(ma[s1],ma[s2]),add(ma[s2],ma[s1]);
		}
		dfs(1);
		if(dp[1][0]>dp[1][1]){
			cout<<dp[1][0]<<" ";
			if(f[1][0]==1) cout<<"Yes"<<endl;
			else cout<<"No"<<endl; 
		}
		else if(dp[1][0]<dp[1][1]){
			cout<<dp[1][1]<<" ";
			if(f[1][1]==1) cout<<"Yes"<<endl;
			else cout<<"No"<<endl;
		}
		else{
			cout<<dp[1][1]<<" "<<"No"<<endl;
		}
	}
	return 0;
}