HDU5536

【题意】给出序列s[], 求max{(s[i]+s[j])^s[k]}，其中i!=j!=k。

【分析】首先看下数据感觉N^3会爆炸，所以想到用字典序优化，枚举s[i]和s[j]，那么从高位往低位查找和s[i]+s[j]尽量每位取反即可，对于三个位置不同，可以先插入所有序列，用到两个删除，用完恢复即可。这样复杂度是O(log2(s[i])*N^2)

【AC CODE】3120ms

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define rep(i,a,n) for(int i = a; i < n; i++)
#define repe(i,a,n) for(int i = a; i <= n; i++)
#define per(i,n,a) for(int i = n; i >= a; i--)
#define clc(a,b) memset(a,b,sizeof(a))
const int INF = 0x3f3f3f3f, MAXN = 1000+10,MAXM = 30;
int a[MAXN],node[MAXN*MAXM][2],vis[MAXN*MAXM],cnt;

void insert(int v, int add)
{
	int u = 0;
	per(i,MAXM-1,0)
	{
		int p = (bool)(v&(1<<i));
		if(-1 == node[u][p])
			node[u][p] = ++cnt;
		u = node[u][p];
		vis[u] += add;
	}
}
int query(int v)
{
	int ans = 0,u = 0;
	per(i,MAXM-1,0)
	{
		int p = (bool)(v&(1<<i));
		if(~node[u][p^1] && vis[node[u][p^1]])
			u = node[u][p^1], ans = (ans<<1)+(p^1);
		else if(~node[u][p] && vis[node[u][p]])
			u = node[u][p], ans = (ans<<1)+p;
		else
			ans = ans<<1;
	}
	return ans;
}
int main()
{
#ifdef SHY
	freopen("d:\\1.txt", "r", stdin);
#endif
	int t;
	scanf("%d", &t);
	while(t--)
	{
		int n;
		scanf("%d", &n);
		cnt = 0;
		clc(node,-1);
		clc(vis,0);
		rep(i,0,n) scanf("%d", &a[i]),insert(a[i],1);
		int ans = 0;
		rep(i,0,n)
		{
			rep(j,i+1,n)
			{
				insert(a[i],-1);
				insert(a[j],-1);
				ans = max(ans,(a[i]+a[j])^query(a[i]+a[j]));
				insert(a[i],1);
				insert(a[j],1);
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}