Werewolf

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1675 Accepted Submission(s): 473

Problem Description

"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≤10

1≤N≤100,000

1≤x≤N

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

Source

2018 Multi-University Training Contest 6

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【总结】

规律题咯。。没智商咯

【题意】

有n个人玩狼人杀。每个人指认一个其他人是狼或者平民，平民只说真话，而狼人不一定。

问最后有多少人必须为平民，多少人必为狼人？

【分析】

首先假设所有都是狼人，是成立的，所以必为平民的人就是0个咯。

一个人若必为狼人，则只有一种情况：1指认2是平民，2指认3是平民.....3指认1是狼，那1就必为狼，还有所有直接或间接指认1是平民的人也必为狼。

也就是说，我只关心指认平民的那个图，一个个的连通块，然后枚举指认别人是狼的那些人，比如u指认了x是狼，如果x直接或间接指认了u是平民，那所有直接或间接指认x和x都是狼。如图：红圈圈起来的都为狼。

那我们用并查集维护平民边构成连通块，遍历说别人是狼人的人，统计如上图这样的红圈中的点数即可。

【代码】

/****
***author: winter2121
****/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX=2e5+5;
struct node{
    int t,next;
}edge[MAX];
int head[MAX],cnt;
void init(int n)
{
    cnt=0; memset(head,-1,sizeof(head[0])*(n+1));
}
void addedge(int u,int v)
{
    edge[cnt]=node{v,head[u]};
    head[u]=cnt++;
}
typedef pair<int,int>PII;
int fa[MAX];
int father(int x){return x==fa[x]?x:fa[x]=father(fa[x]);}
void join(int x,int y){x=father(x);y=father(y);fa[x]=y;}
void dfs(int u,int &ans)
{
    ans++;
    for(int i=head[u];~i;i=edge[i].next)
        dfs(edge[i].t,ans);
}
int main()
{
    int T,n,x,y;
    char s[20];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)fa[i]=i;
        init(n);
        queue<PII>q;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%s",&x,s);
            if(s[0]=='w')q.push(PII(i,x));
            else
            {
                addedge(x,i); //反向边
                join(x,i);
            }
        }
        int ans=0;
        while(!q.empty())
        {
            PII u=q.front(); q.pop();
            if(father(u.first)!=father(u.second))continue;
            //u.second子树全为狼
            dfs(u.second,ans);
        }
        printf("0 %d\n",ans);
    }
    return 0;
}

HDU6370多校6-I（并查集+dfs）

Werewolf

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