Vasiliy's Multiset
题目链接：
http://codeforces.com/contest/706/problem/D
Description
```
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.
</big>
##Input
<big>
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.
</big>
##Output
<big>
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.
</big>
##Examples
<big>
input
10
- 8
- 9
- 11
- 6
- 1
? 3
- 8
? 3
? 8
? 11
output
11
10
14
13
</big>
##Source
<big>
Codeforces Round #367 (Div. 2)
</big>
<br/>
##题意：
<big>
维护一个multiset：
操作1：插入一个数字
操作2：删除指定数字
操作3：给出一个x，求集合中的一个元素y，使得 x XOR y 最大.
</big>
<br/>
##题解：
<big>
由于异或前后的值不能保持单调性，所以排序和线段树都不好处理.
这里需要用一个 0/1-Trie树 来维护.
对于集合中的数用二叉树中的结点来表示其二进制位.
①插入：开辟一些新的树结点.
②删除：找到对应结点赋成NULL. 删除时有几点需要注意：
集合允许重复，所以可以记录每个数出现的次数，当集合中仅有一个x时才删除.
把最底端的叶节点删除后，要递归更新其父结点的状态，否则在匹配时可能会进入不存在的结点.
③匹配：先把x按位取反，从高位到低位在二叉树中匹配x，按能匹配时尽量匹配的原则往下走. 这样可以保证结果从高位到低位有尽量多的位不相同，即异或结果最大.
注意：初始时集合中有个0，一开始忘了这个，RE了2发.
</big>
<br/>
##代码：
``` cpp
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <list>
#define LL long long
#define eps 1e-8
#define maxn 101000
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;
class Trie
{
public:
Trie *next[2];
Trie() {
memset(next,NULL,sizeof(next));
}
};
Trie *root;
void Insert(LL n)
{
Trie *p = root;
for(int i=31;i>=0;i--)
{
int id = (n >> i) & 1;
if(p->next[id] == NULL)
p->next[id] = new Trie();
p = p->next[id];
}
}
int Delete(Trie *p, LL n, int i)
{
if(i < 0) return 1;
int id = (n >> i) & 1;
int ret = Delete(p->next[id], n, i-1);
if(ret) p->next[id] = NULL;
if(p->next[0] == NULL && p->next[1] == NULL)
return 1;
return 0;
}
LL Match(LL m)
{
m = ~m;
LL ans = 0;
Trie *p = root;
for(int i=31;i>=0;i--) {
ans <<= 1;
int bit = (m >> i) & 1;
if(bit) {
if(p->next[1]) {
p = p->next[1];
ans++;
}
else {
p = p->next[0];
}
}
else {
if(p->next[0]) {
p = p->next[0];
}
else {
p = p->next[1];
ans++;
}
}
}
return ans;
}
map<LL, int> occ;
int main(int argc, char const *argv[])
{
//IN;
int q;
while(scanf("%d", &q) != EOF)
{
root = new Trie();
occ.clear();
Insert(0LL);
occ[0] = 1;
while(q--)
{
getchar();
char op; LL val;
op = getchar();
scanf("%I64d", &val);
if(op == '+') {
if(!occ[val]) {
occ[val] = 1;
Insert(val);
}
else occ[val]++;
}
else if(op == '-') {
if(occ[val] == 1) Delete(root, val, 31);
occ[val]--;
} else {
printf("%I64d\n", Match(val)^val);
}
}
}
return 0;
}