bzoj 3689: 异或之 Trie+堆

题目大意:

http://www.lydsy.com/JudgeOnline/problem.php?id=3689

题解:

利用一个优先队列存储当前取到的数
然后再写一颗支持查找异或的k大值的Trie即可
由于同一个值\(x\)可能被\(a_i\text{ xor }a_j\)和\(a_j\text{ xor }a_i\)一起取到
所以只有在奇数次取值的时候再更新

#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
	x=0;char ch;bool flag = false;
	while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
	while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const int maxn = 100010;
struct Node{
	Node *ch[2];
	int siz;
}*null,*root;
Node mem[maxn*33],*it;
inline void init(){
	it = mem;null = it++;
	null->ch[0] = null->ch[1] = null;
	null->siz = 0;root = null;
}
inline Node* newNode(){
	Node *p = it++;p->ch[0] = p->ch[1] = null;
	p->siz = 0;return p;
}
void insert(int x){
	Node *nw = root;nw->siz ++ ;
	for(int i=30;i>=0;--i){
		int id = (x>>i)&1;
		if(nw->ch[id] == null) nw->ch[id] = newNode();
		nw = nw->ch[id];nw->siz ++ ;
	}
}
int kth(int x,int k){
	int ret = 0;Node *nw = root;
	for(int i=30;i>=0;--i){
		int id = (x>>i)&1;
		if(k <= nw->ch[id]->siz){
			nw = nw->ch[id];
		}else{
			k -= nw->ch[id]->siz;
			nw = nw->ch[id^1];
			ret |= (1<<i);
		}
	}
	return ret;
}
int a[maxn];
struct num{
	int val,k,pos;
	bool friend operator < (const num &a,const num &b){
		return a.val > b.val;
	}
	num(const int &a,const int &b,const int &c){
		val = a;k = b;pos = c;
	}
};
priority_queue<num>q;
int main(){
	init();root = newNode();
	int n,k;read(n);read(k);
	for(int i=1;i<=n;++i){
		read(a[i]);
		insert(a[i]);
	}
	for(int i=1;i<=n;++i){
		q.push(num(kth(a[i],2),2,i));
	}
	int cnt = 0;k<<=1;
	while(k--){
		num tp = q.top();q.pop();
		if((++cnt)&1) printf("%d ",tp.val);
		if(tp.k == n) continue;
		q.push(num(kth(a[tp.pos],tp.k+1),tp.k+1,tp.pos));
	}
	getchar();getchar();
	return 0;
}