CF #367 D. Vasiliy's Multiset

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 
     
       Copy 
     
10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11

      output 
     
       Copy 
     
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：n,操作的次数，+ x集合中插入一个元素x；- x集合中删除一个元素x；？ x求集合元素中和x异或的最大值是多少

思路：一道01字典树的题。将要插入的数的二进制位倒着建树(为什么？因为异或时高位尽量大，结果才尽量大)，即高位在深度低的节点上。用一个数组记录经过各个节点的数的个数，插入时，每经过一个点，将节点的这个值加一，删除时，则减一。查找时，当前节点的这个值大于0，说明有数经过。对于要查找的这个数的高位，如果是1，要使异或值尽量大，那么就要往0的地方走，反之，往1的地方走，实在没办法走，只有按原路径走啦。详见代码。注意：0永远在树中。

ac代码:

#include<bits/stdc++.h>
using namespace std;
#define Memset(x, a) memset(x, a, sizeof(x))
typedef long long ll;
const int maxn = 222222;//集合中的数字个数
int ch[32*maxn][2];         //节点的边信息
ll val[32*maxn];            //节点存储的值
int sz,q;                     //树中当前节点个数
ll num;

void init(){
    Memset(ch[0],0);           //树清空
    sz=1;
}

void _insert(ll a){
    int u=0;
    for(int i=32;i>=0;i--){
        int c=((a>>i)&1);
        if(!ch[u][c]){
            Memset(ch[sz],0);
            ch[u][c]=sz++;
        }
        u=ch[u][c];
        ++val[u];
    }
}

void _delete(ll a){
  int u=0;
  for (int i=32; i>=0; --i){
      int c=((a>>i)&1);
      u=ch[u][c];
      --val[u];
  }
}

ll query(ll a){
    int u=0;
    ll ans=0;
    for(int i=32;i>=0;i--){
        int c=((a>>i)&1);
        if (c==1){
            if (ch[u][0] && val[ch[u][0]]){
                ans+=1<<i;
                u=ch[u][0];
            }
            else
                u=ch[u][1];
        }
        else{
            if (ch[u][1] && val[ch[u][1]]){
                ans+=1<<i;
                u=ch[u][1];
            }
            else
                u=ch[u][0];
        }
    }
    return ans;
}

int main()
{
  string op;
  init();
  cin>>q;
  _insert(0);
  while(q--)
    {
    cin>>op>>num;
    if(op[0]=='+')
        _insert(num);
    else if(op[0]=='-')
        _delete(num);
    else
        cout<<query(num)<<endl;
  }
  return 0;
}

CF #367 D. Vasiliy's Multiset

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