"The Werewolves" is a popular card game among young people.In the basic game, there are 2 different groups: the werewolves and the villagers.
Each player will debate a player they think is a werewolf or not.
Their words are like "Player x is a werewolf." or "Player x is a villager.".
What we know is :
1. Villager won't lie.
2. Werewolf may lie.
Of cause we only consider those situations which obey the two rules above.
It is guaranteed that input data exist at least one situation which obey the two rules above.
Now we can judge every player into 3 types :
1. A player which can only be villager among all situations,
2. A player which can only be werewolf among all situations.
3. A player which can be villager among some situations, while can be werewolf in others situations.
You just need to print out the number of type-1 players and the number of type-2 players.
No player will talk about himself.
Input
The first line of the input gives the number of test cases T.Then T test cases follow.
The first line of each test case contains an integer N,indicating the number of players.
Then follows N lines,i-th line contains an integer x and a string S,indicating the i-th players tell you,"Player x is a S."
limits:
1≤T≤101≤T≤10
1≤N≤100,0001≤N≤100,000
1≤x≤N1≤x≤N
S∈S∈ {"villager"."werewolf"}
Output
For each test case,print the number of type-1 players and the number of type-2 players in one line, separated by white space.
Sample Input
1 2 2 werewolf 1 werewolf
Sample Output
0 0
思路：需要发现，只有找到一个环并且一个环中有且只有一个人被指认为狼，其他人都被指认为村民时，才能确定一只狼，当确定一只狼以后，只要说它是村民的，必然也是狼。
所以对于指认是狼的边，直接连，指认是村民的边，连接后并查集合并（这样就可以省去dfs找环）
最后bfs找其他狼就行。
代码：
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>
#include <stack>
using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
#define inf 0x3f3f3f3f
int num;
int head[maxn];
bool vis[maxn];
int f[maxn];
int n;
int Find(int x)
{
if(f[x]==x) return x;
else return f[x]=Find(f[x]);
}
void Union(int x,int y)
{
int fx=Find(x),fy=Find(y);
if(fx==fy) return;
f[fy]=fx;
}
struct Edge
{
int u,v,next,w;
}edge[maxn<<1];
void addEdge(int u,int v,int w)
{
edge[num].u=u;
edge[num].v=v;
edge[num].w=w;
edge[num].next=head[u];
head[u]=num++;
}
void init()
{
memset(head,-1,sizeof(head));
num=0;
}
int main(int argc, char const *argv[])
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
int T;
cin>>T;
while(T--)
{
init();
scanf("%d",&n);
for(int i=1;i<=n;i++) f[i]=i;
for(int i=1;i<=n;i++)
{
int x;
char s[20];
scanf("%d%s",&x,s);
if(s[0]=='w')
{
addEdge(x,i,1);
}
else
{
addEdge(x,i,0);
Union(i,x);
}
}
queue<int> que;
memset(vis,0,sizeof(vis));
int ans=0;
for(int i=0;i<=num;i++)
{
int u=edge[i].u,v=edge[i].v,w=edge[i].w;
if(Find(u)==Find(v)&&w==1&&vis[v]==0)
{
ans++;
que.push(u);
vis[u]=1;
}
}
while(que.empty()==0)
{
int u=que.front();
que.pop();
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v,w=edge[i].w;
if(w) continue;
if(vis[v]) continue;
ans++;
que.push(v);
}
}
printf("0 %d\n",ans);
}
return 0;
}