cf - 706D Vasiliy's Multiset（01字典树）

最新推荐文章于 2022-09-24 16:54:33 发布

Timeclimber

最新推荐文章于 2022-09-24 16:54:33 发布

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分类专栏： Trie字典树文章标签：字典树

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本文介绍了一个关于多重集的算法问题，主要涉及三种操作：添加元素、删除元素和查询最大异或值。通过使用01字典树实现高效的查询和更新过程。

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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
inputCopy
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：

一共有三种操作，+为往多重集里面加入x，-为从多重集里面删除x，？则是从多重集中选出一个数来与x异或使得值最大。

思路：

用01字典树，从高位往地位存入二进制。注意要记录下每个节点出现的次数，因为还有删除操作。

异或的时候取和x当前位的相反就行了，这样才能保证异或出来的最大。

坑点：0一直在多重集里，所以每次都加入0，则一定保证0在多重集合里面了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=2e5+5;
int cnt,root;
struct node
{
    int next[2];
    int num;
}q[40*maxn];
void in(int root,int n)
{
    for(int i=30;i>=0;i--)
    {
        int x=(n>>i)&1;
        if(q[root].next[x]==0)
        {
            q[root].next[x]=++cnt;
            q[q[root].next[x]].num++;
        }
        else
        {
            q[q[root].next[x]].num++;
        }
        root=q[root].next[x];
    }
}
void del(int root,int n)
{
    for(int i=30;i>=0;i--)
    {
        int x=(n>>i)&1;
        if(q[q[root].next[x]].num>0)
        q[q[root].next[x]].num--;
        root=q[root].next[x];
    }
}
int query(int root,int n)
{
    int ans=0;
    for(int i=30;i>=0;i--)
    {
        ans<<=1;
        int x=(n>>i)&1;
        if(q[q[root].next[!x]].num!=0)
        {
            ans=ans|(!x);
            root=q[root].next[!x];
        }
        else
        {
            ans=ans|x;
            root=q[root].next[x];
        }
    }
    return ans^n;
}
int main()
{
    int t,x;
    char op[5];
    while(~scanf("%d",&t))
    {
        cnt=0,root=0;
        memset(q,0,sizeof(q));
        for(int i=1;i<=t;i++)
        {
            scanf("%s",op);
            scanf("%d",&x);
            in(root,0);
            if(op[0]=='+')
            {
                in(root,x);
            }
            else if(op[0]=='-')
            {
                del(root,x);
            }
            else
            {
                printf("%d\n",query(root,x));
            }
        }
    }
    return 0;
}