Codeforces Round #367 (Div. 2) D Vasiliy's Multiset（查找树）

Author has gone out of the stories about Vasiliy, so here is just a formal task description.

You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:

1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value , i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.

Multiset is a set, where equal elements are allowed.

Input

The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.

Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.

Note, that the integer 0 will always be present in the set A.

Output

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.

但只有0和1的时候就是一棵二叉树，然后就是贪心了，尽量找与当前值不一样的

AC代码：

# include <iostream># include <stdio.h># include <stdlib.h># include <malloc.h># include <string>using namespace std;struct node{int cnt;node *zero, *one;};node *newnode();void Insert(int x);void Delete(int x);void Query(int x);node *root=newnode();int flage=0;int main(){int n, i, x;string s;cin>>n;    Insert(0);for(i=1; i<=n; i++){cin>>s>>x;if(s[0]=='+'){Insert(x);flage++;}else if(s[0]=='-'){flage--;Delete(x);}else{Query(x);}}return 0;}node *newnode(){//构造节点 node *p=(node*)malloc(sizeof(node));p->one=NULL;p->zero=NULL;p->cnt=0;return p;}void Insert(int x){//插入 node *p;p=root;for(int i=30; i>=0; i--){int num=(1<<i)&x;if(num){if(p->one==NULL){p->one=newnode();p->one->cnt++;p=p->one;}else{p->one->cnt++;p=p->one;}}else{if(p->zero==NULL){p->zero=newnode();p->zero->cnt++;p=p->zero;}else{p->zero->cnt++;p=p->zero;}}}}void Delete(int x){//删除node *p=root;for(int i=30; i>=0; i--){int num=(1<<i)&x;if(num){p->one->cnt--;p=p->one;}else{p->zero->cnt--;p=p->zero;}}}void Query(int x){//查询 if(!flage){cout<<x<<'\n';return;}node *p=root;int ans=0;for(int i=30; i>=0; i--){int num=(1<<i)&x;if(num){if(p->zero!=NULL&&p->zero->cnt!=0){p=p->zero;ans=ans+(1<<i);}else{p=p->one;}}else{if(p->one!=NULL&&p->one->cnt!=0){p=p->one;ans=ans+(1<<i);}else{p=p->zero;}}}cout<<ans<<'\n';}