Codeforces Round #367 (Div. 2) D. Vasiliy's Multiset （字典树二进制）

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：有三种操作
+ x：添加元素x
− x ：删除元素x
? x：查找元素x和与其异或值最大的数

思路：用字典树按高位存储数字，然后从最高位开始搜索，注意数字x也可以不异或，所以说，每次要先添加一个元素0。

ac代码：

/* ***********************************************Author       : AnICoo1Created Time : 2016-08-19-21.51 FridayFile Name    : D:\MyCode\2016-8月\2016-8-19.cppLANGUAGE     : C++Copyright  2016 clh All Rights Reserved************************************************ */#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<map>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x,y) memset(x,(y),sizeof(x))#define PI acos(-1)#define gn (sqrt(5.0)+1)/2#define eps 1e-8using namespace std;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}//headstruct s{    int num;    s *next[2];};s *root;void create(int x){    s *p=root,*q;    for(int i=30;i>=0;i--)    {        int id=(x>>i)&1;        if(p->next[id]==NULL)        {            q=(s *)malloc(sizeof(s));            q->num=1;            for(int j=0;j<2;j++)            q->next[j]=NULL;            p->next[id]=q;            p=p->next[id];        }        else        {            p->next[id]->num++;            p=p->next[id];        }    }}void Delete(int x){    s *p=root,*q;    for(int i=30;i>=0;i--)    {        int id=(x>>i)&1;        p=p->next[id];        p->num--;    }}int Find(int x){    int ans=0;    s *p=root;    create(0);    for(int i=30;i>=0;i--)    {        int id=(x>>i)&1;        int to=id?0:1;        if(p->next[to]!=NULL&&p->next[to]->num>0)        {            p=p->next[to];            ans=ans*2+to;        }        else        {            p=p->next[id];            ans=ans*2+id;        }    }    return ans;}void Begin(){    for(int i=0;i<2;i++)    root->next[i]=NULL;    root->num=0;}void freetree(s *t){    if(t==NULL)    return;    for(int i=0;i<2;i++)    {        if(t->next[i]!=NULL)        freetree(t->next[i]);    }    free(t);    return;}int main(){    int n;scanf("%d",&n);    root=(s *)malloc(sizeof(s));    Begin();    for(int i=1;i<=n;i++)    {        char ch[2];int x;scanf("%s%d",ch,&x);        if(ch[0]=='+')            create(x);        else if(ch[0]=='-')            Delete(x);        else if(ch[0]=='?')        {            int y=Find(x);            printf("%d\n",x^y);        }    }    freetree(root);    return 0;}