bzoj 4260: REBXOR Trie+乱搞

题目大意:

http://www.lydsy.com/JudgeOnline/problem.php?id=4260

题解:

啊啊啊.

被这种SB题坑了半天.

求出异或前缀和后

从n到1枚举\(r_1\)的取值就好了啊.

用Trie 算出1 ~ \(r_1-1\)和\(a_{r_1}\)的异或最大值ans1

以及\(a_{r_1 + 1}\)和\(r_1 + 1\) ~ n的异或最大值ans2

用ans1 + max{ans2}更新答案就好了

#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 = 600010;
struct Node{
	Node* ch[2];
	int siz;
}*null,*root[maxn];
Node mem[maxn*32],*it;
inline void init(){
	it = mem;
	null = it++;null->ch[0] = null->ch[1] = null;
	null->siz = 0;
}
Node* insert(Node *rt,int x,int d){
	Node *p = it++;*p = *rt;p->siz ++;
	if(d == -1) return p;
	int id = (x>>d)&1;
	p->ch[id] = insert(p->ch[id],x,d-1);
	return p;
}
int query(Node *p1,Node *p2,int x,int d){
	if(d == -1) return 0;
	int id = (x>>d)&1;
	if(p2->ch[id^1]->siz - p1->ch[id^1]->siz > 0)
		return (1<<d) | query(p1->ch[id^1],p2->ch[id^1],x,d-1);
	return query(p1->ch[id],p2->ch[id],x,d-1);
}
void dfs(Node *p){
	printf("siz = %d\n",p->siz);
	if(p->ch[0] != null){puts("-> ch[0] : ");dfs(p->ch[0]);}
	if(p->ch[1] != null){puts("-> ch[1] : ");dfs(p->ch[1]);}
	puts("return");
}
int a[maxn];
int main(){
	int n;read(n);init();root[0] = null;
	root[1] = insert(root[0],0,31);
	++n;
	for(int i=2;i<=n;++i){
		read(a[i]);a[i] ^= a[i-1];
		root[i] = insert(root[i-1],a[i],31);
	}
	int ans = 0,max2 = 0;
	for(int i=n;i>=2;--i){
		max2 = max(max2,query(root[i],root[n],a[i],31));
		ans = max(ans,max2 + query(root[0],root[i],a[i],31));
	}printf("%d\n",ans);
	getchar();getchar();
	return 0;
}