codeforces 842D Trie 树异或不存在的最小值

D. Vitya and Strange Lesson
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Today at the lesson Vitya learned a very interesting function — mex. Mex of a sequence of numbers is the minimum non-negative number that is not present in the sequence as element. For example, mex([4, 33, 0, 1, 1, 5]) = 2 and mex([1, 2, 3]) = 0.

Vitya quickly understood all tasks of the teacher, but can you do the same?

You are given an array consisting of n non-negative integers, and m queries. Each query is characterized by one number x and consists of the following consecutive steps:

Perform the bitwise addition operation modulo 2 (xor) of each array element with the number x.
Find mex of the resulting array.
Note that after each query the array changes.

Input
First line contains two integer numbers n and m (1 ≤ n, m ≤ 3·105) — number of elements in array and number of queries.

Next line contains n integer numbers ai (0 ≤ ai ≤ 3·105) — elements of then array.

Each of next m lines contains query — one integer number x (0 ≤ x ≤ 3·105).

Output
For each query print the answer on a separate line.

Examples
input
2 2
1 3
1
3
output
1
0
input
4 3
0 1 5 6
1
2
4
output
2
0
0
input
5 4
0 1 5 6 7
1
1
4
5
output
2
2
0
2

题意：n个数，m个询问，问异或后不存在的最小的数

逆向思维
每一个数异异或一个数的值都是不一样的，第一种想法是把不存在的数加到字典树上，然后找异或到的最小值,得开 21*（1<<20)的内存有点多了。。

#include <bits/stdc++.h>using namespace std;const int N = (1<<20);const int BB=22;struct node{  int nxt[2];}T[N*BB+1];int mp[(1<<20)];typedef long long ll;int sz;void insert(int d){    int p=0;    for(int i=21;i>=0;i--)    {       int dd=(d>>i)&1;       if(!T[p].nxt[dd]) T[p].nxt[dd]=++sz;       p=T[p].nxt[dd];    }}int go[32];int ans;void ask(){   int p=0;   for(int i=21;i>=0;i--)   {      int dd=go[i];      if(T[p].nxt[dd]) p=T[p].nxt[dd];      else      {           p=T[p].nxt[1-dd],ans|=(1<<i);      }   }}int main(){      memset(T,0,sizeof(T));      int n,m;      scanf("%d%d",&n,&m);      sz=0;      for(int i=1;i<=n;i++)      {            int d;            scanf("%d",&d);            mp[d]=1;      }       for(int i=0;i<=(1<<20);i++)      {          if(!mp[i])          insert(i);      }      int t=0;      while(m--){         int x;         scanf("%d",&x);         t^=x;         ans=0;         for(int j=21;j>=0;j--)         {            int dd=(t>>j)&1;            go[j]=dd;         }         ask();         printf("%d\n",ans);      }}

第二种方法是判断子树是不是满的
往不满的子树走，异或同方向的优先
重复的点要删除

#include <bits/stdc++.h>using namespace std;const int N = 300010;const int BB=22;struct node{  int nxt[2];}T[N*BB+1];int size[N*BB+1];typedef long long ll;int sz;int go[32];int ask(){   int p=0,ans=0;   for(int i=21;i>=0;i--)   {      int dd=go[i];      if(!T[p].nxt[dd]) return ans;      if(size[T[p].nxt[dd]]<(1<<i)) p=T[p].nxt[dd];      else       {          ans|=(1<<i);          p=T[p].nxt[1-dd];          if(!p) return ans;      }   }   return ans;}void insert(int d){    int p=0;    for(int i=21;i>=0;i--)    {       int dd=(d>>i)&1;       if(!T[p].nxt[dd]) T[p].nxt[dd]=++sz;       p=T[p].nxt[dd];       size[p]++;    }    if(size[p]>1)    {        p=0;        for(int i=21;i>=0;i--)        {            int dd=(d>>i)&1;            p=T[p].nxt[dd];            size[p]--;        }    }}int main(){      memset(T,0,sizeof(T));      memset(size,0,sizeof(size));      int n,m;      scanf("%d%d",&n,&m);      sz=0;      for(int i=1;i<=n;i++)      {            int d;            scanf("%d",&d);            insert(d);      }       int tt=0;      while(m--){         int x;         scanf("%d",&x);         tt^=x;         for(int j=21;j>=0;j--)         {            int dd=(tt>>j)&1;            go[j]=dd;         }         printf("%d\n",ask());      }}