D. Vasiliy's Multiset
4 seconds
256 megabytes
standard input
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.
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.
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.
10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11
11
10
14
13
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
.
得学姐指导----涉及到XOR的建树都建字典树。
哇哈哈~
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 5e6+;
struct node
{
int next[];
int v;
};
node tree[maxn];
int sz = ;
void build(int x,int v)
{
int root = ;
for(int i=;i>=;i--)
{
int id = (x>>i)&;
if(tree[root].next[id]==)
{
memset(tree[sz].next,,sizeof(tree[sz].next));
tree[sz].v = ;
tree[root].next[id] = sz++;
}
root = tree[root].next[id];
tree[root].v+=v;
}
}
void match(int x)
{
int root = ;
x = ~x;
int ans = ;
for(int i=;i>=;i--)
{
ans *= ;
int id = (x>>i)&;
if(tree[root].next[id]&&tree[tree[root].next[id]].v)
{
ans++;
root = tree[root].next[id];
}
else
{
root = tree[root].next[-id];
}
}
printf("%d\n",ans);
}
int main()
{
int n,x;
char s[];
cin>>n; /* for(int i=0;i<=maxn-1;i++)
{
tree[i].v = 0;
memset(tree[i].next,0,sizeof(tree[i].next));
}*/
build(,);
for(int i=;i<=n;i++)
{
scanf("%s %d",s,&x);
if(s[]=='+')
{
build(x,);
}
else if(s[]=='-')
{
build(x,-);
}
else
{
match(x);
}
}
return ;
}