Codeforces 706d字典树+贪心

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：multiset中有一个初始的0，三种操作，加入一个元素，删除一个元素，询问一个元素和multiset中一个元素异或的最大值。

题解：字典树，从高位向低位贪心即可，不过注意要打时间戳，直接建树会爆。

#include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int ch[3200000][2]; int n,x,tot=1,ans,sum[3200000]; char s; void add(int x){ int now=1; for(int i=30;i>=0;i--){ if(x&(1<<i)){ if(!ch[now][1])ch[now][1]=++tot;//时间戳 now=ch[now][1]; sum[now]++; } else{ if(!ch[now][0])ch[now][0]=++tot; now=ch[now][0]; sum[now]++; } } } void del(int x){ int now=1; for(int i=30;i>=0;i--){ if(x&(1<<i)){ now=ch[now][1]; sum[now]--; } else{ now=ch[now][0]; sum[now]--; } } } void query(int x) { int now=1; for(int i=30;i>=0;i--){//从高位向低位贪心 if(x&(1<<i)){ if(sum[ch[now][0]]){ ans+=(1<<i); now=ch[now][0]; } else now=ch[now][1]; } else{ if(sum[ch[now][1]]){ ans+=(1<<i); now=ch[now][1]; } else now=ch[now][0]; } } } int main() { cin>>n; add(0); while(n--){ cin>>s>>x; if(s=='+')add(x); if(s=='-')del(x); if(s=='?'){ ans=0; query(x); printf("%d\n",ans); } } return 0; }