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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#602596#8720. BFS 序 0Kanate#RE 0ms22056kbC++142.4kb2024-10-01 11:05:122024-10-01 11:05:12

Judging History

你现在查看的是最新测评结果

  • [2024-10-01 11:05:12]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:22056kb
  • [2024-10-01 11:05:12]
  • 提交

answer

#include<bits/stdc++.h>
#define int long long
//#define mod 1000000007
#define rep(i,j,k) for(int i=(j);i<=(k);i++)
#define per(i,j,k) for(int i=(j);i>=(k);i--)
using namespace std;
template<class T>void chkmax(T &a,T b){a=max(a,b);}
template<class T>void chkmin(T &a,T b){a=min(a,b);}
template<class T>T read(T &x)
{
	x=0;T f=1;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+(c^'0');c=getchar();}
	return x*=f;
}
template<class T,class ...L>void read(T &x,L &...l){read(x),read(l...);}
template<class T>void write(T x)
{
	if(x<0){putchar('-');x=-x;}
	if(x>9){write(x/10);}putchar(x%10+'0');
}
template<class T>void write(T x,char c){write(x),putchar(c);}
int dfnn;
int n,dfn[300005],dep[300005],siz[300005];
int q,m,a[300005];
vector <int> g[300005];
void dfs(int u)
{
	dfn[u]=++dfnn;
	siz[u]=1;
	for(int v:g[u])
	{
		dep[v]=dep[u]+1;
		dfs(v);
		siz[u]+=siz[v];
	}
}
#define mid (l+r)/2
#define ls (o*2)
#define rs (o*2+1)
int seg1[300005<<2],seg2[300005<<2],tag[300005<<2];
void pushup(int o)
{
	seg1[o]=max(seg1[ls],seg1[rs]);
	seg2[o]=min(seg2[ls],seg2[rs]);
}
void pushdown(int o)
{
	if(!tag[o])return;
	tag[ls]=tag[rs]=tag[o];
	seg1[ls]=seg2[ls]=seg1[rs]=seg2[rs]=tag[o]-1;
	tag[o]=0;
	pushup(o);
}
void assign(int lx,int rx,int k,int o=1,int l=1,int r=n)
{
	pushdown(o);
	if(lx<=l&&r<=rx)
	{
		tag[o]=k;
		seg1[o]=k-1;
		seg2[o]=k-1;
		return;
	}
	if(lx<=mid)assign(lx,rx,k,ls,l,mid);
	if(rx>mid)assign(lx,rx,k,rs,mid+1,r);
	pushup(o);
}
int query1(int lx,int rx,int o=1,int l=1,int r=n)
{
	pushdown(o);
	if(lx<=l&&r<=rx)return seg1[o];
	int ans=0;
	if(lx<=mid)chkmax(ans,query1(lx,rx,ls,l,mid));
	if(rx>mid)chkmax(ans,query1(lx,rx,rs,mid+1,r));
	return ans;
}
int query2(int lx,int rx,int o=1,int l=1,int r=n)
{
	pushdown(o);
	if(lx<=l&&r<=rx)return seg2[o];
	int ans=1;
	if(lx<=mid)chkmin(ans,query2(lx,rx,ls,l,mid));
	if(rx>mid)chkmin(ans,query2(lx,rx,rs,mid+1,r));
	return ans;
}

void solve()
{
	assign(1,n,1);
	read(m);
	rep(i,1,m)read(a[i]);
	rep(i,1,m-1)
		if(dep[a[i]]>dep[a[i+1]])
			{puts("No");return;}
	rep(i,1,m)
	{
		int l=dfn[a[i]];
		int r=dfn[a[i]]+siz[a[i]]-1;
		int q1=query1(l,r);
		int q2=query2(l,r);
		if(q1!=q2){puts("No");return;}
		assign(l,r,2);
	}
	puts("Yes");
}
signed main()
{
	read(n);
	rep(i,2,n)
	{
		int fa;read(fa);
		g[fa].push_back(i);
	}
	dfs(1);
	read(q);
	rep(i,1,q)solve();
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 22056kb

input:

6
1 1 3 2 4
10
4 3 6 2 5
1 4
3 2 4 5
5 2 5 4 6 3
3 1 4 2
3 5 6 3
5 4 5 2 6 1
4 4 3 2 5
4 4 6 2 3
3 3 2 6

output:

No
Yes
Yes
No
No
No
No
No
No
Yes

result:

ok 10 token(s): yes count is 3, no count is 7

Test #2:

score: -100
Runtime Error

input:

300000
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1...

output:


result: