codeforces 706D D. Vasiliy's Multiset(trie树)
题目链接:
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
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> #include <map> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar()); for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + '0'); putchar('\n'); } const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=1e18; const int N=2e5+10; const int maxn=5e3+4; const double eps=1e-12; char s[100]; int x; int ch[32*N][2],sz=1,val[32*N][2],temp[50]; void in(int num) { int u=0; for(int i=31;i>=0;i--) { int c=temp[i];val[u][c]+=num; if(!ch[u][c]) { ch[sz][0]=ch[sz][1]=0; ch[u][c]=sz++; } u=ch[u][c]; } } int query() { int u=0,ans=0; for(int i=31;i>=0;i--) { int c=temp[i]; if(ch[u][c^1]&&val[u][c^1]) { ans|=(1<<i); u=ch[u][c^1]; } else u=ch[u][c]; } return ans; } int main() { mst(temp,0); in(1); int q; read(q); while(q--) { scanf("%s%d",s,&x); mst(temp,0); int cnt=0; while(x) { temp[cnt++]=x%2; x>>=1; } if(s[0]=='+')in(1); else if(s[0]=='-')in(-1); else printf("%d\n",query()); } return 0; }