codeforces 367 D. Vasiliy's Multiset
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
- "+ 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
- "? 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
Copy
10 + 8 + 9 + 11 + 6 + 1 ? 3 - 8 ? 3 ? 8 ? 11
output
Copy
11 10 14 13
題意:初始有個一個空集,n此操作,操作分三種
1 + x,將一個x加入集合
2 - x,刪除集合內的一個x
3 ?x,詢問集合中與x異或的最大值
求集合內異或最大值一般用字典樹來求,所以建立一個字典樹維護一下。
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct node
{
int cnt;
node *next[2];
node()
{
cnt = 0;
memset(next, 0, sizeof(next));
}
};
node *root = new node();
void insert(int x)
{
node *p = root;
for (int i = 30;i >= 0;i--)
{
int num = x&(1 << i) ? 1 : 0;
if (p->next[num] == NULL)
p->next[num] = new node();
p = p->next[num];
p->cnt++;
}
}
void del(int x)
{
node *p = root;
for (int i = 30;i >= 0;i--)
{
int num = x&(1 << i) ? 1 : 0;
p = p->next[num];
p->cnt--;
}
}
int query(int x)
{
node *p = root;
int ans = 0;
for (int i = 30;i >= 0;i--)
{
int num = x&(1 << i) ? 0 : 1;
if (p->next[num] != NULL&&p->next[num]->cnt > 0)
ans += 1 << i, p = p->next[num];
else p = p->next[num ^ 1];
}
return ans;
}
int main()
{
int n, i, x;
char str[2];
insert(0);
scanf("%d", &n);
for (i = 1;i <= n;i++)
{
scanf("%s%d", str, &x);
if (str[0] == '+')
insert(x);
else if (str[0] == '-')
del(x);
else
printf("%d\n", query(x));
}
return 0;
}