ACM学习历程—HDU 5536 Chip Factory（xor && 字典树）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <queue>
#include <vector>
#define LL long long

using namespace std;

//求n个数中取出两个数能xor的最大值
//Trie Tree字典树
//len*n, len为数的二进制最大长度
const int maxN = 1005;
const int len = 31;//len表示数的二进制最大长度
struct Trie
{
    int next[2];
}tree[maxN*len];
int num[maxN*len];
int cnt, n, a[maxN];

void initTree()
{
    cnt = 0;
    memset(tree, -1, sizeof(tree));
    memset(num, 0, sizeof(num));
}

void add(int x)
{
    int now = 0;
    bool k;
    for (int i = len; i >= 0; i--)
    {
        k = x&(1<<i);
        if (tree[now].next[k] == -1)
            tree[now].next[k] = ++cnt;
        now = tree[now].next[k];
        num[now]++;
    }
}

void del(int x)
{
    int now = 0;
    bool k;
    for (int i = len; i >= 0; i--)
    {
        k = x&(1<<i);
        now = tree[now].next[k];
        num[now]--;
    }
}

//返回当前数中能和x合成最大数的数
int query(int x)
{
    int v = 0, now = 0;
    bool k;
    for (int i = len; i >= 0; i--)
    {
        k = x&(1<<i);
        if (tree[now].next[!k] != -1 && num[tree[now].next[!k]] != 0)
            k = !k;
        v = v|(k<<i);
        now = tree[now].next[k];
    }
    return v;
}

void input()
{
    initTree();
    scanf("%d", &n);
    for (int i = 0; i < n; ++i)
    {
        scanf("%d", &a[i]);
        add(a[i]);
    }
}

int myMax(int x, int y)
{
    if (x == -1) return y;
    else return max(x, y);
}

void work()
{
    int ans = -1, tmp;
    for (int i = 0; i < n; ++i)
    {
        for (int j = i+1; j < n; ++j)
        {
            del(a[i]); del(a[j]);
            tmp = a[i]+a[j];
            //cout << "****" << tmp << " " << query(tmp) << endl;
            ans = myMax(ans, tmp^query(tmp));
            add(a[i]); add(a[j]);
        }
    }
    printf("%d\n", ans);
}

int main()
{
    //freopen("test.in", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int times = 0; times < T; ++times)
    {
        input();
        work();
    }
    return 0;
}