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

最新推荐文章于 2022-12-16 16:19:21 发布

say_c_box

最新推荐文章于 2022-12-16 16:19:21 发布

阅读量913

点赞数

CC 4.0 BY-SA版权

分类专栏：算法数据结构字符串文章标签：算法 codeforces 数据结构

本文链接：https://blog.youkuaiyun.com/say_c_box/article/details/52190895

算法同时被 3 个专栏收录

246 篇文章

订阅专栏

暑假训练+个人复习

84 篇文章

订阅专栏

数据结构

46 篇文章

订阅专栏

本文解析了一个涉及多集合操作的问题，包括添加、删除元素及查询最大异或值的操作，并通过字典树实现高效处理大量查询。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 $\text{[math]}$ , 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 x_i (1 ≤ x_i ≤ 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 x_i 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 $\text{[math]}$ — maximum among integers $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ and $\text{[math]}$ .

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.youkuaiyun.com/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;
}