标签：ring redo fir nes ems before turn element property
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:
, 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 
.
就是有3种操作
+ x向集合里添加 x
- x 删除x元素,(保证存在
? x 查询 x | 集合中元素的最大值
新姿势 get，就是利用字典树，从高位到低位进行贪心。想到从高位求，但是不知道怎么用字典树..orz.新姿势
1 #include <algorithm> 2 #include <stack> 3 #include <istream> 4 #include <stdio.h> 5 #include <map> 6 #include <math.h> 7 #include <vector> 8 #include <iostream> 9 #include <queue> 10 #include <string.h> 11 #include <set> 12 #include <cstdio> 13 #define FR(i,n) for(int i=0;i<n;i++) 14 #define MAX 2005 15 #define mkp pair <int,int> 16 using namespace std; 17 //#pragma comment(linker, "/STACK:10240000000,10240000000") 18 const int maxn = 4e6+4; 19 typedef long long ll; 20 const int inf = 0x3fffff; 21 void read(int &x) { 22 char ch; bool flag = 0; 23 for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == ‘-‘)) || 1); ch = getchar()); 24 for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar()); 25 x *= 1 - 2 * flag; 26 } 27 28 int tree[maxn][3]; 29 int sum[maxn]; 30 31 int cnt = 2; 32 void Insert(int val,ll t) 33 { 34 int root = 1; 35 for(int i=31;i>=0;i--) 36 { 37 int id=(t>>i)&1; 38 if(!tree[root][id]) 39 { 40 tree[root][id]=cnt++; 41 } 42 root = tree[root][id]; 43 sum[root]+=val; 44 } 45 } 46 47 ll ans(ll tmp) 48 { 49 tmp=~tmp; 50 int root = 1; 51 ll res = 0; 52 for(int i=31;i>=0;i--) 53 { 54 int id = (tmp>>i)&1; 55 res*=2; 56 if(tree[root][id]&&sum[tree[root][id]]) 57 { 58 res++; 59 root = tree[root][id]; 60 } 61 else 62 { 63 root=tree[root][1-id]; 64 } 65 } 66 return res; 67 } 68 69 int main() { 70 int n; 71 read(n); 72 memset(sum,0,sizeof(sum)); 73 memset(tree,0,sizeof(tree)); 74 Insert(1,0); 75 /*for(int i=0;i<32;i++){ 76 printf("%d %d\n",tree[i][0],tree[i][1]); 77 }*/ 78 for(int i=0;i<n;i++){ 79 char c; 80 ll t; 81 cin>>c>>t; 82 if(c==‘+‘)Insert(1,t); 83 else if(c==‘-‘)Insert(-1,t); 84 else { 85 printf("%lld\n",ans(t)); 86 } 87 } 88 return 0; 89 }
标签：ring redo fir nes ems before turn element property
原文地址：https://www.cnblogs.com/DreamKill/p/9427765.html
