hihocoder1860 最大异或和

思路：

把N个前缀异或和插入一棵trie树中，然后对每个前缀异或和x计算能使x ^ y最大的前缀异或和y。利用了异或运算的a ^ b ^ a = b的性质。

参考了https://cloud.tencent.com/developer/article/1343206

实现：

 1 #include <iostream>
 2 using namespace std;
 3 
 4 const int INF = 0x3f3f3f3f;
 5 const int N = 100005;
 6 int a[N], cnt = 0;
 7 
 8 struct trieNode
 9 {
10     trieNode * next[2];
11 }; 
12 
13 trieNode pool[N * 32];
14 
15 trieNode * createTrie()
16 {
17     trieNode * root = &pool[cnt++];
18     for (int i = 0; i < 2; i++)
19     {
20         root->next[i] = NULL;    
21     }
22     return root;
23 } 
24 
25 void insert(trieNode * root, int x)
26 {
27     trieNode * tmp = root;
28     int index;
29     for (int i = 31; i >= 0; i--)
30     {
31         int msk = (1 << i);
32         if (msk & x) index = 1;
33         else index = 0;
34         if (tmp->next[index] == NULL)
35         {
36             tmp->next[index] = createTrie();
37         }
38         tmp = tmp->next[index];
39     }
40 }
41 
42 int search(trieNode * root, int x)
43 {
44     trieNode * tmp = root;
45     int res = 0;
46     for (int i = 31; i >= 0; i--)
47     {
48         int msk = 1 << i;
49         int need;
50         if (msk & x) need = 0;
51         else need = 1;
52         if (i == 31) need = 1 - need;
53         if (tmp->next[need])
54         {
55             res += (need << i);
56             tmp = tmp->next[need];
57         }
58         else
59         {
60             res += (1 - need << i);
61             tmp = tmp->next[1 - need];
62         }
63     }
64     return x ^ res;
65 }
66 
67 int main()
68 {
69     int n;
70     while (cin >> n)
71     {
72         cnt = 0;
73         trieNode * root = createTrie();
74         for (int i = 1; i <= n; i++)
75         {
76             cin >> a[i];
77             a[i] = a[i - 1] ^ a[i];
78             insert(root, a[i]);
79         }
80         insert(root, 0);
81         int ans = -INF;
82         for (int i = 1; i <= n; i++)
83         {
84             ans = max(ans, search(root, a[i]));
85         }
86         cout << ans << endl;
87     }
88     return 0;
89 }