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

say_c_box發表於2016-08-12

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:

  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.

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。

以前都不知道這種用字典樹搞。

關於字典樹的入門可以看這個   http://blog.csdn.net/say_c_box/article/details/52073513

查詢操作就是，找到可以讓該位為1的兒子節點，向下訪問。因為越高位是1數就越大嘛。

最多不過30層，所以直接建30層的樹即可。

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;
}








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