CodeForces 367D Vasiliy's Multiset Trie树

D. Vasiliy's Multiset

time limit per test
4 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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:

"+ x" — add integer x to multiset A.

"- 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.

"? 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.

Example

input

10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11

output

11
10
14
13

Note

After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.

The answer for the sixth query is integer 

 —
maximum among integers 

, 

, 

, 

and 

.

题意：你有一个多重集合，一开始只有0，现在有3种操作：1.增加一个元素。2.擦除一个元素(如果有多个只擦除一个)。3给你一个x，问你集合里和他亦或以后最大是多少。

题解：把里面的元素弄成二进制，然后弄一颗Trie树,每次询问的时候从高位贪心的选择就好了。

//************************************************************************//
//*Author : Handsome How                                                 *//
//************************************************************************//
//#pragma comment(linker, "/STA    CK:1024000000,1024000000")
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <string>
#include <ctime>
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define foreach(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define fur(i,a,b) for(int i=(a);i<=(b);i++)
#define furr(i,a,b) for(int i=(a);i>=(b);i--)
#define cl(a) memset((a),0,sizeof(a))
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#ifdef HandsomeHow
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define dbg(x) cout << #x << " = " << x << endl
#else
#define debug(...)
#define dbg(x)
#endif
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair <int, int> pii;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const int mod=1000000007;
const double pi=acos(-1);
inline void gn(long long&x){
int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;
}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
ll gcd(ll a,ll b){return a? gcd(b%a,a):b;}
ll powmod(ll a,ll x,ll mod){ll t=1ll;while(x){if(x&1)t=t*a%mod;a=a*a%mod;x>>=1;}return t;}
// (づ°ω°)づe★
//-----------------------------------------------------------------
struct Node{
int v;
Node* son[2];
}Root,*root;

void add(int x){
Node *p = root;
furr(i,30,0){
int t = (x>>i)&1;
if(p->son[t] == NULL){
p->son[t] = new Node;
p = p->son[t];
p->son[0] = p->son[1] = NULL;
p->v = 1;
}else{
p = p->son[t];
p->v++;
}
}
}

void era(int x){
Node *p = root;
furr(i,30,0){
int t = (x>>i)&1;
p = p->son[t];
p->v--;
}
}

int query(int x){
Node *p = root;
int v = 0;
add(0);
furr(i,30,0){
int t = (x>>i)&1;
if(p->son[!t]!=NULL && p->son[!t]->v>0){
p = p->son[!t];
v = v * 2 + (!t);
}else{
p = p->son[t];
v = v * 2 + t;
}
}
return x^v;
}

char op[10];
int x;
int main(){
#ifdef HandsomeHow
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
time_t beginttt = clock();
#endif
root = &Root;
root->son[0] = root->son[1] = NULL;
root->v = 0;
int q;
scanf("%d",&q);
while(q--){
scanf("%s %d",op,&x);
if(op[0] == '+') add(x);
if(op[0] == '-') era(x);
if(op[0] == '?') printf("%d\n",query(x));
}
#ifdef HandsomeHow
time_t endttt = clock();
debug("time: %d\n",(int)(endttt - beginttt));
#endif
return 0;
}