Codeforces Round #367 (Div. 2) D. Vasiliy's Multiset Trie

题目链接：

http://codeforces.com/contest/706/problem/D

D. Vasiliy's Multiset

time limit per test:4 seconds
memory limit per test:256 megabytes #### 问题描述 > 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.

输入

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.

输出

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.

样例

sample input  
10  
+ 8  
+ 9  
+ 11  
+ 6  
+ 1  
? 3  
- 8  
? 3  
? 8  
? 11  

sample output  
11  
10  
14  
13  

题意

维护一个multiset，支持插入删除，并且对于查询“？ X”输出还在集合中的元素中与X最大的异或和。

题解

用trie树来维护集合，对于x，把它拆成32位二进制，然后按照从高位开始存到Trie里面，对于查询，只要在深搜Trie树的时候从高位开始故意往与x相反的方向跑就可以了。

代码

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=(a);i<(b);i++)

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=1e9;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;

//start----------------------------------------------------------------------

const int maxnode=1e7+10;
const int maxm=34;

int arr[66],tot;
int ch[maxnode][2];
int val[maxnode],tag[maxnode];
struct Trie{
	int sz;
	Trie(){
		sz=1; clr(ch[0],0); 
		clr(val,0);
		clr(tag,-1);
	}
	void insert(int x,int type){
		int tmp=x;
		int u=0, n=maxm;
		tot=0;
		clr(arr,0);
		while(x){ arr[tot++]=x%2; x/=2; }
		for(int i=n-1;i>=0;i--){
			int c=arr[i];
			val[u]+=type;
			if(!ch[u][c]){
				clr(ch[sz],0);
				val[sz]=0;
				ch[u][c]=sz++;
			}
			u=ch[u][c];
		}
		val[u]+=type;
		tag[u]=tmp;
	}
	int query(int x){
		int tmp=x;
		int ret=0;
		int u=0, n=maxm;
		tot=0;
		clr(arr,0);
		while(x){ arr[tot++]=x%2; x/=2; }
		int su=1;
		for(int i=n-1;i>=0;i--){
			int c=arr[i];
//			printf("c:%d
",c);
//			printf("val[%d]:%d
",u,val[u]);
			if(ch[u][c^1]&&val[ch[u][c^1]]>0){
				u=ch[u][c^1];
			}else if(ch[u][c]&&val[ch[u][c]]>0){
				u=ch[u][c];
			} 
			else{
				su=0;
				break;
			}
		} 
		if(su) ret=max(tmp,tag[u]^tmp);
		else ret=tmp;
		return ret;
	}
}trie;

int main() {
	int q;
	scanf("%d",&q);
	while(q--){
		char cmd[11]; int x;
		scanf("%s%d",cmd,&x);
		if(cmd[0]=='+') trie.insert(x,1);
		else if(cmd[0]=='-') trie.insert(x,-1); 
		else if(cmd[0]=='?') printf("%d
",trie.query(x)); 
	}
	return 0;
}

//end-----------------------------------------------------------------------

Notes

对于xor的操作，十有八九是要拆位考虑，拆完之后可以考虑用线段树，Trie树之类的数据结构去维护。