2019 icpc南昌邀请赛 G Winner

题目链接：https://nanti.jisuanke.com/t/40259

Ichuan really likes to play games, so he organized a game competition with NN participating players.

Follows are the rule of the game competition.

1. There are three modes in the game, in each mode, players have different ability values, in addition, each player may have different ability value in different mode.

2. There are a total of N-1 matches. In each match, two players who have not yet been eliminated will play against each other in one of the modes. The player who has high ability in this mode will win, and the other one will be eliminated.

3. The only player who remains in the game after all N-1 matches will be the winner.

As the organizer of the game, ichuan can manipulate the result of the game. Namely, for each match, he can choose both players and match mode. Of course, he can only choose players who have not yet been eliminated.

Ichuan has some friends, some of them will ask him: "Does player XX will be the winner?" Answering this question will give you a lot of reward, so you need to write a program which can answer these questions.

Input

The first line contains integers N and Q(1≤N,Q≤105), the number of players and the number of requests.

The next three lines, each line contains N integers, the Xth​ integer Y(1≤Y≤106)represents the ability value of player X in this mode.,

The next qq lines, each line only has one integer X(1≤X≤n), which means ichuan's friend wants to know if player X can be the winner.

Output

For each query, if player X has a chance to be the winner, output "YES", otherwise output "NO"

输出时每行末尾的多余空格，不影响答案正确性

样例输入复制

4 4 
1 2 3 4
1 2 4 3
2 1 3 4
1 
2 
3
4

样例输出复制

NO
NO 
YES 
YES   

题意：有n个选手，有3个比赛模式，每个模式下每个选手有一个实力值，实力值高者可以打败低者，裁判可以任意选择两个选手在任意模式下比赛，实力高者获胜，实力低着淘汰，q个询问，问裁判是否可以帮助玩家i在经历n-1场获胜。

思路：三个模式，在每个模式下，按照实力排序，再将相邻实力高者与实力低者连边，缩点，入度为0的强连通分量的点都是可以获胜的点。

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const int maxm=1e6+10;
struct node{
    int u,v,w,nxt;
}e[maxm];
int h[maxn],low[maxn],dfn[maxn],vis[maxn];
int st[maxn],belong[maxn],du[maxn];
int n,q,cnt,tot,top,num,y;
struct nd{
    int x,id;
}a[maxn];
bool cmp(nd a1,nd a2)
{
    return a1.x<a2.x; 
}
void init()
{
    memset(h,-1,sizeof(h));
    memset(belong,0,sizeof(belong));
    memset(du,0,sizeof(du));
    memset(dfn,0,sizeof(dfn));
    memset(vis,0,sizeof(vis));
    cnt=tot=num=top=0;
}

void add(int u,int v)
{
    e[cnt].u=u,e[cnt].v=v;
    e[cnt].nxt=h[u];h[u]=cnt++;
}

void tarjan(int u)//缩点 
{
    dfn[u]=low[u]=++tot;
    vis[u]=1;
    st[++top]=u;
    for(int i=h[u];i!=-1;i=e[i].nxt)
    {
        int v=e[i].v;
        if(!dfn[v]) 
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);    
        }
        else if(vis[v])
            low[u]=min(low[u],dfn[v]);
    }
    if(dfn[u]==low[u])
    {
        int t;
        num++;
        do{
            t=st[top--];
            vis[t]=0;
            belong[t]=num;
        }while(t!=u);
    }
}

int main()
{
    init();
    scanf("%d%d",&n,&q);
    for(int i=1;i<=3;i++)
    {
        for(int j=1;j<=n;j++)
        {
            scanf("%d",&a[j].x);
            a[j].id=j;
        }
        sort(a+1,a+n+1,cmp);//排序 
        for(int j=1;j<n;j++)//连边 
            add(a[j+1].id,a[j].id);
    }
    for(int i=1;i<=n;i++)
        if(!dfn[i])
            tarjan(i);
    for(int i=0;i<cnt;i++)//计算入度 
        if(belong[e[i].u]!=belong[e[i].v])
            du[belong[e[i].v]]++;
    for(int i=1;i<=q;i++)
    {
        scanf("%d",&y);
        if(du[belong[y]]==0)
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}