Query on A Tree
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 1692 Accepted Submission(s): 558
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?
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 V1,V2,⋯,Vn, indicating the value of node i.
The second line contains n-1 non-negative integers F1,F2,⋯Fn−1, Fi 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≤n,q≤105
0≤Vi≤109
1≤Fi≤n, the root of the tree is node 1.
1≤u≤n,0≤x≤109
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
Sample Output
2 3
Source
Recommend
liuyiding
题意：给一棵树，树上每个节点有一个权值。然后有q次查询，每次查询给你节点标号u和一个数x。问以u的子树里面的所有节点点权和数x的最大异或值是多少。
解题思路：主要用来学习字典树合并，还是很简单的。合并的过程天然是启发式的。这题用可持久化也能AC，主要是要先把dfs序求出来，然后就像主席树那样操作就好了。字典树合并的话，要注意，由于是启发式的，因此合并的时候，子树的结构会被破坏，因此在合并前，要先把相关的答案计算出来。
可持久化解法：2400ms+
#include<iostream>
#include<string.h>
#include<queue>
#include<bitset>
#include<vector>
using namespace std;
typedef long long ll;
const int MAXN=100005;
const int INF=1e18;
vector<int> G[MAXN];
int root=1;//总的根的个数
int ch[MAXN*33][2];//子节点
int num[MAXN*33];//当前编号的节点有多少个
int tot=0;//总的点数
int rt[MAXN];//第i棵树的根的编号
int a[MAXN];
void insert(int x,int r,int lr){
bitset<32> s(x);
int cur=r;
int lcur=lr;
for(int i=31;i>=0;i--){
if(!ch[cur][s[i]]){
ch[cur][s[i]]=++tot;
ch[cur][!s[i]]=ch[lcur][!s[i]];
num[ch[cur][s[i]]]=num[ch[lcur][s[i]]];
}
cur=ch[cur][s[i]];
lcur=ch[lcur][s[i]];
num[cur]++;
}
}
int lft[MAXN],riht[MAXN];
int temp=0;
void insertdfs(int u,int fa){
rt[u]=++tot;
lft[u]=u;
riht[u]=u;
insert(a[u],rt[u],rt[temp]);
temp=u;
for(int i=0;i<G[u].size();i++){
int v=G[u][i];
if(v!=fa){
insertdfs(v,u);
riht[u]=riht[v];
}
}
}
int query(int x,int rt,int lrt){
bitset<32> s(x);
bitset<32> ans;
int ucur=rt;
int vcur=lrt;
for(int i=31;i>=0;i--){
if(num[ch[vcur][!s[i]]]-num[ch[ucur][!s[i]]]>0){
//printf("%d %d %d\n",i,num[ch[ucur][!s[i]]],num[ch[vcur][!s[i]]]);
ans[i]=1;
ucur=ch[ucur][!s[i]];
vcur=ch[vcur][!s[i]];
}
else{
ucur=ch[ucur][s[i]];
vcur=ch[vcur][s[i]];
}
}
return ans.to_ullong();
}
int main(){
int N,Q;
while(~scanf("%d%d",&N,&Q)){
tot=0;
temp=0;
memset(ch,0,sizeof(ch));
memset(num,0,sizeof(num));
memset(rt,0,sizeof(rt));
for(int i=1;i<=N;i++){
scanf("%d",&a[i]);
G[i].clear();
}
int v;
for(int i=2;i<=N;i++){
scanf("%d",&v);
G[v].push_back(i);
}
insertdfs(1,0);
int u,x;
while(Q--){
scanf("%d%d",&u,&x);
//cout<<lft[u]<<" "<<riht[u]<<endl;
printf("%d\n",max(x^a[u],query(x,rt[lft[u]],rt[riht[u]])));
}
}
return 0;
}
字典树启发式合并解法 1700ms+
#include<iostream>
#include<string.h>
#include<queue>
#include<bitset>
#include<vector>
using namespace std;
typedef long long ll;
const int MAXN=100005;
const int INF=1e18;
vector<int> G[MAXN];
int ch[MAXN*33][2];//子节点
int num[MAXN*33];//当前编号的节点有多少个
int tot=0;//总的点数
int rt[MAXN];//第i棵树的根的编号
int a[MAXN];
vector<pair<int,int > > qy[MAXN];
void insert(int x,int r){
bitset<32> s(x);
int cur=r;
for(int i=31;i>=0;i--){
if(!ch[cur][s[i]]){
ch[cur][s[i]]=++tot;
}
cur=ch[cur][s[i]];
num[cur]++;
}
}
//超简单的
int merge(int rt,int lrt){
if(!rt)return lrt;
if(!lrt)return rt;
ch[rt][0]=merge(ch[rt][0],ch[lrt][0]);
ch[rt][1]=merge(ch[rt][1],ch[lrt][1]);
return rt;
}
int query(int x,int rt){
bitset<32> s(x);
bitset<32> ans;
int ucur=rt;
for(int i=31;i>=0;i--){
if(num[ch[ucur][!s[i]]]>0){
ans[i]=1;
ucur=ch[ucur][!s[i]];
}
else{
ucur=ch[ucur][s[i]];
}
}
return ans.to_ullong();
}
int ans[MAXN];
int temp=0;
void insertdfs(int u,int fa){
rt[u]=++tot;
insert(a[u],rt[u]);
for(int i=0;i<G[u].size();i++){
int v=G[u][i];
if(v!=fa){
insertdfs(v,u);
rt[u]=merge(rt[u],rt[v]);
}
}
//先计算了，以免后面合并时被破坏结构
for(auto &q: qy[u]){
ans[q.first] = query(q.second,rt[u]);
}
}
int main(){
int N,Q;
while(~scanf("%d%d",&N,&Q)){
tot=0;
temp=0;
memset(ch,0,sizeof(ch));
memset(num,0,sizeof(num));
memset(rt,0,sizeof(rt));
for(int i=1;i<=N;i++){
scanf("%d",&a[i]);
G[i].clear();
qy[i].clear();
}
int v;
for(int i=2;i<=N;i++){
scanf("%d",&v);
G[v].push_back(i);
}
int u,x;
for(int i=0;i<Q;i++){
scanf("%d%d",&u,&x);
qy[u].push_back(make_pair(i,x));
}
insertdfs(1,0);
for(int i = 0; i < Q; i++)
printf("%d\n", ans[i]);
}
return 0;
}