cf430(div.2)D Vitya and Strange Lesson 字典树

Description

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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).

Solution

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大意就是给定n个数字，m个询问，每次给出正整数x，求所有数字异或上x后的mex

字典树的一大用处就是兹磁对一些数字进行二进制操作
我们知道所有数异或上x可以用打标记的方法，即令与x异或为1的层交换左右儿子
对于求mex的操作可以贪心走最小。一开始我用size判断是否为满结果用一个bool类表示当前节点的子树是否为满就行了(lll￢ω￢)

Code

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#include <stdio.h>
#include <string.h>
#include <algorithm>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)
#define drp(i,st,ed) for (int i=st;i>=ed;--i)

const int N=800005;

int rec[N][2],tot;
int tag[N];

bool full[N];

int read() {
    int x=0,v=1; char ch=getchar();
    for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):(v),ch=getchar());
    for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
    return x*v;
}

void ins(int &now,int dep,int x) {
    if (!now) now=++tot;
    if (dep==-1) {
        full[now]=1;
        return ;
    }
    int ch=(x&(1<<dep))!=0;
    ins(rec[now][ch],dep-1,x);
    full[now]=full[rec[now][0]]&full[rec[now][1]];
}

void push_down(int now,int pos) {
    if (!now||!tag[now]) return ;
    tag[rec[now][0]]^=tag[now];
    tag[rec[now][1]]^=tag[now];
    if ((1<<pos)&tag[now]) {
        std:: swap(rec[now][0],rec[now][1]);
    }
    tag[now]=0;
}

int mex(int now,int dep,int x) {
    if (dep==-1)  return 0;
    push_down(now,dep);
    if (!full[rec[now][0]]) {
        return mex(rec[now][0],dep-1,x);
    } else {
        return mex(rec[now][1],dep-1,x)+(1<<dep);
    }
}

int main(void) {
    int n=read(),T=read(),root=0;
    rep(i,1,n) ins(root,21,read());
    for (;T--;) {
        int x=read();
        tag[root]^=x;
        int ans=mex(root,21,x);
        printf("%d\n", ans);
    }
    return 0;
}