【JZOJ4811】排队

description

---

analysis

• 堆(+)树上倍增

• 考虑后序遍历搞出(dfs)序，那么要填肯定是从(dfs)序开始填

• 把每个点是序里第几位看成优先级，用小根堆来维护当前空着的优先级最小的点

• 插入每次弹(x)次堆顶，然后把这些点全部打上标记，注意标记一定是先打儿子再打父亲

• 然后找一个点深度最浅的打过标记的祖先，由于标记肯定打完了该点到祖先的所有点，于是倍增就可以找出

• 找完祖先后，把该祖先入堆，删除标记即可

---

code

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<iostream>
#define MAXN 100005
#define MAXM MAXN*2
#define ll long long
#define fo(i,a,b) for (ll i=a;i<=b;++i)
#define fd(i,a,b) for (ll i=a;i>=b;--i)

using namespace std;

vector<ll>a[MAXN];
ll b[MAXN],c[MAXN],f[MAXN],heap[MAXN],depth[MAXN];
ll pre[MAXN][17];
bool bz[MAXN];
ll n,t,tot;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline bool cmp(ll x,ll y){return f[x]<f[y];}
inline void swap(ll &x,ll &y){ll z=x;x=y,y=z;}
inline void dfs(ll x,ll y)
{
	fo(i,0,a[x].size()-1)if (a[x][i]!=y)depth[a[x][i]]=depth[x]+1,dfs(a[x][i],x);
	f[x]=++tot,pre[x][0]=y;
}
inline void insert(ll x)
{
	heap[++heap[0]]=x;ll now=heap[0];
	while (c[heap[now>>1]]>c[heap[now]] && now>1)swap(heap[now],heap[now>>1]),now>>=1;
}
inline ll get()
{
	ll tmp=heap[1],now=1;
	heap[1]=heap[heap[0]--],heap[heap[0]+1]=0;
	while ((now<<1)<=heap[0])
	{
		ll next=now<<1;
		if (next<heap[0] && c[heap[next+1]]<c[heap[next]])++next;
		if (c[heap[now]]<=c[heap[next]])break;
		swap(heap[now],heap[next]),now=next;
	}
	return tmp;
}
inline ll find(ll x)
{
	fd(i,16,0)if (bz[pre[x][i]])x=pre[x][i];
	return x;
}
int main()
{
	freopen("T1.in","r",stdin);
	n=read(),t=read();
	fo(i,1,n-1)
	{
		ll x=read(),y=read();
		a[x].push_back(y);
		a[y].push_back(x);
	}
	fo(i,1,n)sort(a[i].begin(),a[i].end()),b[i]=i;
	tot=0,dfs(1,0),sort(b+1,b+n+1,cmp);
	fo(i,1,n)c[b[i]]=i;
	fo(i,1,n)insert(b[i]);
	fo(j,1,16)fo(i,1,n)pre[i][j]=pre[pre[i][j-1]][j-1];
	while (t--)
	{
		ll op=read(),x=read(),y;
		if (op==1)
		{
			fo(i,1,x)bz[y=get()]=1;
			printf("%lld
",y);
		}
		else
		{
			y=find(x),printf("%lld
",depth[x]-depth[y]);
			insert(y),bz[y]=0;
		}
	}
	return 0;
}