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:

"+ 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

, 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

 — 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 InputRaw

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

Sample OutputRaw

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);
}

}