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DFS重标号+二分

打比赛的时候想了很多方法..DFS序，BFS序，倍增什么的都考虑了一遍，但是几乎要么是可以维护两个区间但是代码复杂度爆炸，要么就是只能维护单一维度的信息。

这道题的具体做法就是先DFS遍历一遍，记一下每个点的出入栈时间。按照每个点的深度排序。这样就可以二分出深度，然后根据入栈的时间可以再在这个深度内二分出合适的区间。范围就是询问节点的出入栈时间。

//CF 570D//by Cydiater//2016.10.14#include <iostream>#include <cstdlib>#include <queue>#include <map>#include <ctime>#include <cmath>#include <map>#include <cstdio>#include <string>#include <cstring>#include <algorithm>using namespace std;#define ll long long#define up(i,j,n)		for(int i=j;i<=n;i++)#define down(i,j,n)		for(int i=j;i>=n;i--)const int MAXN=1e6+5;const int oo=0x3f3f3f3f;inline int read(){	char ch=getchar();int x=0,f=1;	while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}	while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}	return x*f;}int LINK[MAXN],len=0,N,M,tim=0,pos[MAXN],Xor[MAXN],lef1,lef2,rig1,rig2;char s[MAXN];struct edge{	int y,next;}e[MAXN];struct Node{	int in,out,dep,id,v;}a[MAXN];namespace solution{	inline void insert(int x,int y){e[++len].next=LINK[x];LINK[x]=len;e[len].y=y;}	inline bool cmp(Node x,Node y){return x.dep==y.dep?x.in<y.in:x.dep<y.dep;}	void dfs(int Node,int deep){		a[Node].in=++tim;a[Node].dep=deep;a[Node].id=Node;		for(int i=LINK[Node];i;i=e[i].next)dfs(e[i].y,deep+1);		a[Node].out=++tim;	}	void init(){		N=read();M=read();		up(i,2,N){			int Node=read();			if(i==Node)continue;			insert(Node,i);		}		scanf("%s",s+1);		up(i,1,N)a[i].v=s[i]-‘a‘;		dfs(1,1);		sort(a+1,a+N+1,cmp);		up(i,1,N){			pos[a[i].id]=i;			Xor[i]=((1<<a[i].v)^Xor[i-1]);		}	}	int get1(int lim){//<=		int leftt=1,rightt=N,mid;		while(leftt+1<rightt){			mid=(leftt+rightt)>>1;			if(a[mid].dep>=lim)	rightt=mid;			else                leftt=mid;		}		if(a[leftt].dep>=lim)		return leftt;		return rightt;	}	int get2(int lim){//>=		int leftt=1,rightt=N,mid;		while(leftt+1<rightt){			mid=(leftt+rightt)>>1;			if(a[mid].dep<=lim)	leftt=mid;			else                rightt=mid;		}		if(a[rightt].dep<=lim)		return rightt;		return leftt;	}	int get3(int lim){//<=		int leftt=lef1,rightt=rig1,mid;		while(leftt+1<rightt){			mid=(leftt+rightt)>>1;			if(a[mid].in>=lim)	rightt=mid;			else                leftt=mid;		}		if(a[leftt].in>=lim)		return leftt;		return rightt;	}	int get4(int lim){//>=		int leftt=lef1,rightt=rig1,mid;		while(leftt+1<rightt){			mid=(leftt+rightt)>>1;			if(a[mid].in<=lim)	leftt=mid;			else                rightt=mid;		}		if(a[rightt].in<=lim)		return rightt;		return leftt;	}	void slove(){		while(M--){			int Node=pos[read()],dep=read();			if(a[Node].dep>=dep){puts("Yes");continue;}			lef1=get1(dep);rig1=get2(dep);			lef2=get3(a[Node].in+1);rig2=get4(a[Node].out-1);			int S=Xor[rig2]^Xor[lef2-1];int flag=0;			if(lef2>rig2){				puts("Yes");				continue;			}			up(i,0,25)if((S&(1<<i))==(1<<i)&&flag){				puts("No");				flag=-1;				break;			}else if((S&(1<<i))==(1<<i)) flag=1;			if(flag!=-1)puts("Yes");		}	}}int main(){	//freopen("input.in","r",stdin);	using namespace solution;	init();	slove();	return 0;}

CF570D：Tree Requests