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
.
题目链接：CF 706D
字典树水题，只是最近在膜可持久化字典树，拿这道题当模版写了一下，感觉跟主席树非常像，毕竟都是可持久化的思想，新建节点信息代替更改节点信息，写法也简单明了如果会主席树的话代码是不难理解的，只是最后返回的值是异或过的值，因为一个节点不再是模拟地去初始化、连接到链表尾，而是直接把节点的位置拿来标记为已使用，把信息复制到对应节点上去，因此需要一个变量cnt来记录这条路径有没有被覆盖过或者说有没有被使用过，有的话说明这条路到底是存在一个数，即用普通字典树的话来说就是这个点不为NULL，可以往下走。滋瓷范围内的Xor最大值的贪心查询
代码：
#include <bits/stdc++.h>
using namespace std;
const int N=200010;
struct Trie
{
int nxt[2];
int cnt;
};
Trie L[N*34];
int tot;
int root[N]; void init()
{
memset(L,0,sizeof(L));
tot=0;
}
void update(int &cur,int ori,int step,int n,int v)
{
cur=++tot;
L[cur]=L[ori];
L[cur].cnt+=v;
if(step<0)
return ;
int t=(n>>step)&1;
update(L[cur].nxt[t],L[ori].nxt[t],step-1,n,v);
}
int Find(int S,int E,int step,int n)
{
if(step<0)
return 0;
int t=(n>>step)&1;
if(L[L[E].nxt[t^1]].cnt-L[L[S].nxt[t^1]].cnt>0)
return (1<<step)+Find(L[S].nxt[t^1],L[E].nxt[t^1],step-1,n);
else
return Find(L[S].nxt[t],L[E].nxt[t],step-1,n);
}
int main(void)
{
int n,x,i;
char ops[3]; while (~scanf("%d",&n))
{
init();
update(root[1],root[0],31,0,1);
int op=1;
int sz=0;
for (i=1; i<=n; ++i)
{
scanf("%s",ops);
if(ops[0]=='+')
{
++sz;
++op;
scanf("%d",&x);
update(root[op],root[op-1],31,x,1);
}
else if(ops[0]=='-')
{
--sz;
++op;
scanf("%d",&x);
update(root[op],root[op-1],31,x,-1);
}
else
{
scanf("%d",&x);
if(!sz)
{
printf("%d\n",x);
continue;
}
printf("%d\n",Find(root[0],root[op],31,x));
}
}
}
}