NBUT1597：Find MaxXorSum（字典树）

• [1597] Find MaxXorSum

• 时间限制: 2000 ms　内存限制: 65535 K
• 问题描述
• Given n non-negative integers, you need to find two integers a and b that a xor b is maximum. xor is exclusive-or.
• 输入
• Input starts with an integer T(T <= 10) denoting the number of tests.
  For each test case, the first line contains an integer n(n <= 100000), the next line contains a1, a2, a3, ......, an(0 <= ai <= 1000000000);
• 输出
• For each test case, print you answer.
• 样例输入

• 2
  4
  1 2 3 4
  3
  7 12 5

• 样例输出

• 7
  11

• 提示

• 无

• 来源

• Alex@NBUT

题意：给N个数，求其中两个数异或的最大值。

思路：字典树，将N个数的二进制存到树里，然后再遍历N个数，找到异或的最大值。

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# define ll long long
using namespace std;
const int maxn = 1e5;

ll a[maxn+3], cnt=0;
struct node
{
    int next[2];
}tri[maxn*13];//乘以13即可，待证。

void add(ll x)
{
    int cur = 0;
    for(int i=31; i>=0; --i)//题目数据范围为32位整型。
    {
        int t = (x>>i)&1;
        if(!tri[cur].next[t])
            tri[cur].next[t] = ++cnt;
        cur = tri[cur].next[t];
    }
}

ll query(ll x)
{
    int cur = 0;
    ll tmp = 0;
    for(int i=31; i>=0; --i)
    {
        int t = !((x>>i)&1);//找到相反的，因为不同的异或值为1。
        if(tri[cur].next[t])
        {
            tmp |= (1<<i);
            cur = tri[cur].next[t];
        }
        else
            cur = tri[cur].next[!t];
    }
    return tmp;
}

int main()
{
    int t, n;
    scanf("%d",&t);
    while(t--)
    {
        cnt = 0;
        scanf("%d",&n);
        memset(tri, 0, sizeof(tri));
        for(int i=1; i<=n; ++i)
        {
            scanf("%lld",&a[i]);
            add(a[i]);
        }
        ll ans = 0;
        for(int i=1; i<=n; ++i)
            ans = max(ans, query(a[i]));
        printf("%lld\n",ans);
    }
    return 0;
}