AcWing 143. 最大异或对

AcWing 143. 最大异或对

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思路

按照每一位是 0 是 1 建立字典树
由于异或时不同为 1 ，查找时按照反方向查找

注意

1. 如果反方向不存在记得回到正方向
2. 从高位到低位建树，相当于贪心策略，1 在高位更优

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//看题解前的丑代码...
#include <iostream>
#include <cstdio>
const int N = 100005;
int tot = 1;
int trie[N][2], end[N<<5], a[N];
int max(int a, int b)
{
	return a > b ? a : b;
}
void insert(int x)
{
	int p = 1;
	for(int i = 31; i >= 0; i--)
	{
		int next = (a[x] & (1 << i)) >> i;
		if(!trie[p][next]) trie[p][next] = ++tot;
		p = trie[p][next];
	}
	end[p] = x;
}
int search(int x)
{
	int p = 1;
	for(int i = 31; i >= 0; i--)
	{
		int next = (x & (1 << i)) >> i ^ 1;
		if(!trie[p][next]) next = next ^ 1;
		p = trie[p][next];
	}
	return a[end[p]];
}
int main(){
	int n;
	scanf("%d", &n);
	int ans = 0;
	for(int i = 0; i < n; ++i)
	{
		scanf("%d", &a[i]);
		//printf("%d %d %d\n", search(a[i]), a[i], search(a[i]) ^ a[i]);
		ans = max(ans, search(a[i]) ^ a[i]);
		insert(i);
	}
	printf("%d\n", ans);
	return 0;
}

//借鉴后...
#include <iostream>
#include <cstdio>
const int N = 100005;
int tot = 1;
int trie[N<<5][2], a[N];
int max(int a, int b)
{
	return a > b ? a : b;
}
void insert(int x)
{
	int p = 1;
	for(int i = 30; i >= 0; i--)
	{
		int next = (a[x] >> i) & 1;
		if(!trie[p][next]) trie[p][next] = ++tot;
		p = trie[p][next];
	}
}
int search(int x)
{
	int p = 1;
	int ans = 0;
	for(int i = 30; i >= 0; i--)
	{
		int next = (x >> i) & 1;
		if(trie[p][next ^ 1])
		{
			p = trie[p][next ^ 1];
			ans |= 1 << i;
		}
		else
			p = trie[p][next];
	}
	return ans;
}
int main(){
	int n;
	scanf("%d", &n);
	int ans = 0;
	for(int i = 0; i < n; ++i)
	{
		scanf("%d", &a[i]);
		//printf("%d %d %d\n", search(a[i]), a[i], search(a[i]) ^ a[i]);
		ans = max(ans, search(a[i]));
		insert(i);
	}
	printf("%d\n", ans);
	return 0;
}