字典树（Codeforces Round #367 (Div. 2) Vasiliy's Multiset，Xor问题 ）

1.题目链接：http://www.codeforces.com/problemset/problem/706/D

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 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  — maximum among integers , , , and .

 题意：有个多重集合，我们可以往里面加入一个数和删除一个数。也可以询问这个集合中的数与x异或的最最大值。

解法：对于求一个数对集合中的异或值最大，我们可以将集合中的树看成一个由其二进制组成的字符串。高位在前，低位在后。我们要求x与集合异或的最大值，我们只需在字典树中按位查找到与（~x）尽量相匹配的值（即高位尽可能的相等）。

AC：

#include<algorithm>
#include<iostream>
#include<cmath>
#include<map>
#include<string.h>
#include<cstring>
#include<vector>
#include<queue>
#include<stdio.h>
using namespace std;
#define ll  long long int
#define maxn 3000008
const int mod=20071027;
int ch[maxn][2];
int val[maxn];
struct trie
{
    public:

    int sz=1;
    void  tt(){   sz=1;   memset(ch[0],0,sizeof(ch[0]) );  };//初始化
    void insert_(int v,int x)//插入，如果是删除的操作的话权值为-1
    {
        int tp,u=0;
        for(int i=30;i>=0;i--)
        {
             tp=(x>>i)&1;
            if(!ch[u][tp])
            {
                memset(ch[sz],0,sizeof(ch[sz]));
                val[sz]=0;
                ch[u][tp]=sz++;
            }
            u=ch[u][tp];
            val[u]+=v;

        }

    };
   int search_(int x)//查找到与~x尽量匹配的值
    {
        int tp,u=0;
        int ans=0;
        for(int i=30;i>=0;i--)
        {
             tp=1-((x>>i)&1);
             if(!ch[u][tp])//如果不存在者位，则取另一位
            {
                tp=1-tp;
            }
            else if(val[ch[u][tp]]<=0)//下一个的价值为0，不能取。只能取另一位
            {
                tp=1-tp;
            }
            u=ch[u][tp];

          //  printf("%d:%d,",tp,val[u]);
            ans+=tp<<i;
        }

        //cout<<endl;
        return ans;
    }
}t;
int main(void)
{
  //freopen("in.txt","r",stdin);
  int q,x;
  char s[10];
  scanf("%d",&q);
  t.tt();
  t.insert_(1,0);
  while(q--)
  {
      scanf("%s%d",&s,&x);
      if(s[0]=='+')  t.insert_(1,x);
      else if(s[0]=='-')  t.insert_(-1,x);
      else if(s[0]=='?')
      {
          printf("%d\n",x^t.search_(x));
      }
  }

}

题目2：链接： https://www.oj.swust.edu.cn/problem/show/2475

2475: Xor问题

Description

问题很简单，现在有一个数组a1，a2，a3……an。你的任务就是找到一个连续子段[l,r]，使得al^al+1^……^ar达到最大。

Input

多组输入，每组有两行。第一行有一个整数n(1<=n<=10^5)，表示数组的元素个数。第二行有n个元素，依次表示数组的元素。(0<=ai<=10^6)

Output

每组输出一行，这行仅一个数字。表示最大的连续子段异或值。

• Sample Input
• Raw

  5
  1 2 3 4 5
  5
  2 3 2 3 2

• Sample Output
• Raw

  7
  3
  

  
  

  解法：对于任意的L~R区间我们都可以用1->r区间的异或值异或上1->(l-1).所以我们可以将1-i( 0<i<=n)的异或值加入字典树，然后遍历找到其中最大的值既可以。

  AC：

  #include<algorithm>
  #include<iostream>
  #include<cmath>
  #include<map>
  #include<string.h>
  #include<cstring>
  #include<vector>
  #include<queue>
  #include<stdio.h>
  using namespace std;
  #define ll  long long int
  #define maxn 1000008
  const int mod=20071027;
  int ch[maxn][3];
  int a[maxn];
  struct trie
  {
      public:
      int sz=1;
      void  tt(){ sz=1; memset(ch[0],0,sizeof(ch[0]) );  };
      void insert_(int x)
      {
          int tp,u=0;
          for(int i=25;i>=0;i--)
          {
               tp=(x>>i)&1;
              if(!ch[u][tp])
              {
                  memset(ch[sz],0,sizeof(ch[sz]));
                  ch[u][tp]=sz++;
              }
              u=ch[u][tp];
          }
      };
     int search_(int x)
      {
          int tp,u=0;
          int ans=0;
          for(int i=25;i>=0;i--)
          {
               tp=1-((x>>i)&1);
               if(!ch[u][tp])
              {
                  tp=1-tp;
              }
              u=ch[u][tp];
            //  printf("%d:%d,",tp,val[u]);
              ans+=tp<<i;
          }
   
          //cout<<endl;
          return x^ans;
      }
  }t;
  int main(void)
  {
      //freopen("in.txt","r",stdin);
       int n;
       a[0]=0;
       int k=0;
       while( scanf("%d",&n)!=EOF)
      {
           t.tt();
           t.insert_(0);
           for( int j=1; j<=n; j++)
          {
              int d;
              scanf("%d",&a[j]);
              a[j]^=a[j-1];
              t.insert_(a[j]);
          }
          int ans=0;
          for(  int j=1;j<=n;j++)
          {
              ans=max(ans,t.search_(a[j]));
          }
          printf("%d\n",ans);
      }
   
   
   
  }