Codeforces 706d 字典树+贪心

最新推荐文章于 2022-04-01 12:31:11 发布

black_miracle

最新推荐文章于 2022-04-01 12:31:11 发布

阅读量461

点赞数

CC 4.0 BY-SA版权

分类专栏：贪心字典树

本文链接：https://blog.youkuaiyun.com/black_miracle/article/details/52202381

贪心同时被 2 个专栏收录

16 篇文章

订阅专栏

字典树

3 篇文章

订阅专栏

本题涉及处理一系列关于整数集合的操作，包括添加、删除和查询最大异或值。采用字典树实现高效处理，适用于算法竞赛。

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 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 $\text{[math]}$ — maximum among integers $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ and $\text{[math]}$ .

题意：multiset中有一个初始的0，三种操作，加入一个元素，删除一个元素，询问一个元素和multiset中一个元素异或的最大值。

题解：字典树，从高位向低位贪心即可，不过注意要打时间戳，直接建树会爆。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int ch[3200000][2];
int n,x,tot=1,ans,sum[3200000];
char s;
void add(int x){
    int now=1;
    for(int i=30;i>=0;i--){
        if(x&(1<<i)){ 
            if(!ch[now][1])ch[now][1]=++tot;//时间戳 
            now=ch[now][1];
            sum[now]++;
        }
        else{
            if(!ch[now][0])ch[now][0]=++tot;
            now=ch[now][0];
            sum[now]++;
        }
    }
}
void del(int x){
    int now=1;
    for(int i=30;i>=0;i--){
        if(x&(1<<i)){
            now=ch[now][1];
			sum[now]--;
        }
        else{
            now=ch[now][0];
			sum[now]--;
        }
    }
}
void query(int x)
{
    int now=1;
    for(int i=30;i>=0;i--){//从高位向低位贪心
        if(x&(1<<i)){
            if(sum[ch[now][0]]){
                ans+=(1<<i);
                now=ch[now][0];
            }
            else now=ch[now][1];
        }
        else{
            if(sum[ch[now][1]]){
                ans+=(1<<i);
                now=ch[now][1];
            }
            else now=ch[now][0];
        }
    }
}
int main()
{
    cin>>n;
    add(0);
    while(n--){
        cin>>s>>x;
        if(s=='+')add(x);
        if(s=='-')del(x);
        if(s=='?'){
            ans=0;
            query(x);
            printf("%d\n",ans);
        }
    }
    return 0;
}