Query on A Tree
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 1238 Accepted Submission(s): 444
Problem Description
Monkey A lives on a tree, he always plays on this tree.
One day, monkey A learned about one of the bit-operations, xor. He was keen of this interesting operation and wanted to practise it at once.
Monkey A gave a value to each node on the tree. And he was curious about a problem.
The problem is how large the xor result of number x and one node value of label y can be, when giving you a non-negative integer x and a node label u indicates that node y is in the subtree whose root is u(y can be equal to u).
Can you help him?
One day, monkey A learned about one of the bit-operations, xor. He was keen of this interesting operation and wanted to practise it at once.
Monkey A gave a value to each node on the tree. And he was curious about a problem.
The problem is how large the xor result of number x and one node value of label y can be, when giving you a non-negative integer x and a node label u indicates that node y is in the subtree whose root is u(y can be equal to u).
Can you help him?
Input
There are no more than 6 test cases.
For each test case there are two positive integers n and q, indicate that the tree has n nodes and you need to answer q queries.
Then two lines follow.
The first line contains n non-negative integers $V_1,V_2,cdots,V_n$, indicating the value of node i.
The second line contains n-1 non-negative integers $F_1,F_2,cdots\,F_{n-1}$, $F_i$ means the father of node $i+1$.
And then q lines follow.
In the i-th line, there are two integers u and x, indicating that the node you pick should be in the subtree of u, and x has been described in the problem.
$2 leq n,q leq 10^5$
$0 leq V_i leq 10^9$
$1 leq F_i leq n$, the root of the tree is node 1.
$1 leq u leq n,0 leq x leq 10^9$
For each test case there are two positive integers n and q, indicate that the tree has n nodes and you need to answer q queries.
Then two lines follow.
The first line contains n non-negative integers $V_1,V_2,cdots,V_n$, indicating the value of node i.
The second line contains n-1 non-negative integers $F_1,F_2,cdots\,F_{n-1}$, $F_i$ means the father of node $i+1$.
And then q lines follow.
In the i-th line, there are two integers u and x, indicating that the node you pick should be in the subtree of u, and x has been described in the problem.
$2 leq n,q leq 10^5$
$0 leq V_i leq 10^9$
$1 leq F_i leq n$, the root of the tree is node 1.
$1 leq u leq n,0 leq x leq 10^9$
Output
For each query, just print an integer in a line indicating the largest result.
Sample Input
2 2 1 2 1 1 3 2 1
很久都没写过博客了qwq，重新写起了久违的博客。
其实前几天才做了18年的湖北省赛的网络同步赛，I题完全就是这个题的弱化版，而且数据还很水，被我写错了一个地方还AC了。
经人提点后，才发现对于一颗树上的每一个节点以及他的子节点经过DFS序可以构成一个连续的区间（转换成区间后就完全就是跟湖北省赛的I一模一样的问题，实在是无力吐槽），但是有一个需要注意的地方，dfs序的标号与原先的节点标号是不一样的，所以需要用一个数组记录他们之间的映射关系，然后通过离线即可解决问题。
关于离线的具体操作，就是将所有的询问保存下来，然后按照左端点的位置从大到小排序，然后从大到小插入大于每个询问值对应的二进制到01字典树中，同时更新或者标记对应点的id，如果查询时遇到id的值小于等于询问的右区间即可取得区间异或的最大值，否则反之。
#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <list>
#include <map>
#include <new>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
using namespace std;
# ifdef __GNUC__
# if __cplusplus > 199711L
# include <unordered_set>
# include <unordered_map>
# else
#include <tr1/unordered_map>
#include <tr1/unordered_set>
using tr1::unordered_map;
using tr1::unordered_set;
# endif
# else
# include <unordered_map>
# include <unordered_set>
# endif
#define FOR(i,a,n) for(register int i=(a),_=(n)+1;i<_;++i)
#define FIR(i,n,a) for(register int i=(n),_=(a)-1;i>_;--i)
#define all(a) (a).begin(),(a).end()
#define vlb(a,n) (lower_bound(all(a),n)-(a).begin())
#define vub(a,n) (upper_bound(all(a),n)-(a).begin())
#define vlbx(a,n) ((a)[(lower_bound(all(a),n)-(a).begin())])
#define vubx(a,n) ((a)[(upper_bound(all(a),n)-(a).begin())])
#define reunique(a) (a).resize(unique(all(a))-(a).begin())
#define mem(a,b) memset((a),(b),sizeof(a))
#define sz(x) (int((x).size()))
#define lowbit(x) ((x)&(-x))
#define lch p<<1,l,mid
#define rch p<<1|1,mid+1,r
#define ll (p<<1)
#define rr (p<<1|1)
#define queues priority_queue
#define pb push_back
#define mp(a,b) make_pair((a),(b))
#define lb lower_bound
#define ub upper_bound
#define ff first
#define ss second
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//__attribute__((optimize("-O2")))
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef vector<bool> VB;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<PII> VII;
typedef vector<PLL> VLL;
// STL other **************************************************************************************
template<class T,class U>inline istream& operator >> (istream& os,pair<T,U>& p) { return os>>p.first>>p.second; }
template<class T,class U>inline pair<T,U> operator + (const pair<T,U>& p1,const pair<T,U>& p2) { return mp(p1.ff+p2.ff,p1.ss+p2.ss); }
template<class T,class U>inline pair<T,U>& operator += (pair<T,U>& p1,const pair<T,U>& p2) { p1.ff+=p2.ff,p1.ss+=p2.ss; return p1; }
template<class T,class U>inline pair<T,U> operator - (const pair<T,U>& p1,const pair<T,U>& p2) { return mp(p1.ff-p2.ff,p1.ss-p2.ss); }
template<class T,class U>inline pair<T,U>& operator -= (pair<T,U>& p1,const pair<T,U>& p2) { p1.ff-=p2.ff,p1.ss-=p2.ss; return p1; }
// Useful constants *******************************************************************************
const int primes[7] = {24443, 100271, 1000003, 1000333, 5000321, 98765431, 1000000123};
const int dx[]= { 0, 1, 0,-1, 0, 1,-1, 1,-1};
const int dy[]= {-1, 0, 1, 0,-1, 1,-1,-1, 1};
#define ee 2.718281828459
#define eps 0.00000001
#define fftmod 998244353
#define INF 0x3f3f3f3f
#define LINF 0xfcfcfcfcfcfcfcfll
#define MOD 1000000007
#define pi 3.14159265358979323846l
int teble();
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
#ifndef ONLINE_JUDGE
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#endif
teble();
return 0;
}
//***************************************************************
//***************************************************************
//***************************************************************
/**________________________code.begin_________________________**/
const int N=1e5+5;
int a[N],b[N],tot,ans[N];
PII p[N];
struct Edge_Node {
int head[N],cnt;
struct Edge { int to,next; } edge[N];
void init() { mem(head,0),cnt=1; }
void add(int u,int v) {
edge[cnt]={v,head[u]},head[u]=cnt++;
}
} E;
struct Dictionarie_Tree {
int tree[N<<5][3],top;
void init() { mem(tree[0],0),top=1; }
void ins(int x,int id) {
int rt,nxt,u,t;
for(rt=0,t=31;~t;rt=nxt,--t) {
nxt=tree[rt][u=x>>t&1];
if(!nxt)mem(tree[top],0),tree[rt][u]=nxt=top++;
tree[nxt][2]=id;
}
}
int query(int x,int l) {
int rt,nxt,u,t,ans=0;
for(rt=0,t=31;~t;rt=nxt,--t) {
u=x>>t&1,ans<<=1;
if((nxt=tree[rt][!u])&&tree[nxt][2]<=l)ans+=1;
else nxt=tree[rt][u];
}
return ans;
}
} D;
struct Node {
int id,x,l,r;
bool operator < (Node &m) const {
return l>m.l;
}
} node[N];
void init() {
E.init(),D.init(),tot=0;
}
void dfs(int x) {
p[x].ff=++tot,b[tot]=x;
for(int i=E.head[x];i;i=E.edge[i].next) {
dfs(E.edge[i].to);
}
p[x].ss=tot;
}
int teble() {
int n,q,x,y;
while(cin>>n>>q) {
init();
FOR(i,1,n)cin>>a[i];
FOR(i,2,n)cin>>x,E.add(x,i);
dfs(1);
FOR(i,1,q)cin>>x>>y,node[i]={i,y,p[x].ff,p[x].ss};
sort(node+1,node+n+1);
int s=n;
FOR(i,1,q) {
while(s>=node[i].l)D.ins(a[b[s]],s),--s;
ans[node[i].id]=D.query(node[i].x,node[i].r);
}
FOR(i,1,q)cout<<ans[i]<<endl;
}
return 0;
}
/**_________________________code.end__________________________**/
//***************************************************************
//***************************************************************
//***************************************************************
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