Codeforces Round #367 (Div. 2) D——Vasiliy's Multiset（异或字典树）

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 

.

新技能get啊。

以前都不知道这种用字典树搞。

关于字典树的入门可以看这个   http://blog.csdn.net/say_c_box/article/details/52073513

查询操作就是，找到可以让该位为1的儿子节点，向下访问。因为越高位是1数就越大嘛。

最多不过30层，所以直接建30层的树即可。

有一句一定要注意：Note, that the integer 0 will
always be present in the set A.

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
using namespace std;
const int MAXN =100000+10;
const long long INF =100000000000007;

struct node{
node *next[2];
int cnt;
node(){
memset(next,NULL,sizeof(next));
cnt=0;
}
};
node *p,*root=new node();
void _insert(int x){
p=root;
for(int i=30;i>=0;i--){
int num=x&(1<<i)?1:0;
if(p->next[num]==NULL)
p->next[num]=new node();
p=p->next[num];
p->cnt++;
}
}

void _delete(int x){
p=root;
for(int i=30;i>=0;i--){
int num=x&(1<<i)?1:0;
p=p->next[num];
p->cnt--;
}
}

int query(int x){
int res=0;
p=root;
for(int i=30;i>=0;i--){
int num=x&(1<<i)?0:1;
node *temp;
temp=p->next[num];
if(temp&&temp->cnt>0){
res+=pow(2.0,i);
p=temp;
}
else{
p=p->next[!num];
}
}
return res;
}

int main(){
int q;
_insert(0);
scanf("%d",&q);
getchar();
while(q--){
char op[2];
int x;
scanf("%s%d",op,&x);
if(op[0]=='+')
_insert(x);
else if(op[0]=='-')
_delete(x);
else
printf("%d\n",query(x));
}
return 0;
}