ACM: Find MaxXorSum 解题报告-字典树

Find MaxXorSum
Time Limit:2000MS     Memory Limit:65535KB     64bit IO Format:

Description
Given n non-negative integers, you need to find two integers a and b that a xor b is maximum. xor is exclusive-or.
Input
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);
Output
For each test case, print you answer.
Sample Input
2
4
1 2 3 4
3
7 12 5

Sample Output
7
11

这个题目其实思路很简单，就我知道的写法有Trie树指针和静态数组两种写法，一开始写的指针用了pow函数TLE了2次，然后换位运算又TLE两次醉了。。后面换成静态数组数组开大了RE了一次。。。哭。。。这是AC代码：

 1 #include"iostream"
 2 #include"algorithm"
 3 #include"cstdio"
 4 #include"cmath"
 5 #include"cstring"
 6 #define MX 1400000
 7 #define INF 0x3f3f3f3f
 8 #define lson l,m,rt<<1
 9 #define rson m+1,r,rt<<1|1
10 using namespace std;
11 int tree[MX][2],xx;
12
13 void BuildTrie(long long a) {
14     int i=31;
15     int p=0;
16     while(0<=i) {
17         bool num=a&(1<<i);
18         if(!tree[p][num]) {        //如果节点为空
19             tree[p][num]=++xx;//标记并创建新的子节点
20         }
21         p=tree[p][num];
22         //    cout<<"YES"<<i<<"   B is:"<<num<<"\n"; //建立的Trie树的样子
23         i--;
24     }
25 }
26
27 long long Query(long long  a) {
28     int i=31,p=0;
29     long long ans=0;
30     while(0<=i) {
31         bool num=!(a&(1<<i));  //取反查找
32         if(tree[p][num]) p=tree[p][num],ans+=1<<i;//如果和原来的那一为相反的存在的话，返回值就加上，并且在这个支路走
33         else p = tree[p][!num];  //按照相同的顺序找
34     //    cout<<"YES"<<i<<"   B is:"<<num<<"     ans is:"<<ans<<endl;   //查询时候的Trie树的样子
35         i--;
36     }
37     return ans;
38 }
39
40 int main() {
41     int T,n;
42     long long a[100050],ans;
43     scanf("%d",&T);
44     while(T--) {
45         ans=0;
46         memset(tree,0,sizeof(tree));
47         xx=0;
48         scanf("%d",&n);
49         for(int i=1; i<=n; i++) {
50             scanf("%I64d",&a[i]);
51             BuildTrie(a[i]);
52         }
53         //cout<<"YES1\n";
54         for(int i=1; i<=n; i++) {
55             ans=max(ans,Query(a[i]));
56         }
57         //    cout<<"YES2\n";
58         printf("%I64d\n",ans);
59     }
60     return 0;
61 }

这是指针的写法，感觉复杂度和上面的没啥区别，为啥就ＴＬＥ了呢？　希望有人知道给我指点下。

 1 #include"iostream"
 2 #include"algorithm"
 3 #include"cstdio"
 4 #include"cmath"
 5 #include"cstring"
 6 #define MX 110000
 7 #define INF 0x3f3f3f3f
 8 #define lson l,m,rt<<1
 9 #define rson m+1,r,rt<<1|1
10 using namespace std;
11
12 struct Trie {
13     Trie *next[2];
14 } root;
15
16 void BuildTrie(int a) {
17     Trie *p=&root,*q;
18     int i=31;
19     while(i>=0) {
20     bool num=a&(1<<i);
21         if(p->next[num]==NULL) {
22             q=(Trie *)malloc(sizeof(root));//申请一块新内存; //动态分配内存
23             q->next[1]=NULL;
24             q->next[0]=NULL;    //清空申请内存的所有子节点
25             p->next[num]=q;        //往子节点下去继续记录字典树
26             p=p->next[num];
27         } else {
28             p=p->next[num];
29         }
30     //    cout<<"YES"<<i<<"   B is:"<<num<<"\n"; //建立的Trie树的样子
31         i--;
32     }
33 }
34
35 long long Query(int a) {
36     Trie *p=&root;
37     long long ans=0;
38     int i=31;
39     while(0<=i) {
40         bool num=a&(1<<i);
41         if(p->next[!num]!=NULL) p=p->next[!num],ans+=1<<(i);//如果和原来的那一为相反的存在的话，返回值就加上，并且在这个支路走
42         else if(p->next[num]!=NULL) p=p->next[num];  //按照相同的顺序找
43     //    cout<<"YES"<<i<<"   B is:"<<num<<"\n"<<"ans is:"<<ans<<endl;   //查询时候的Trie树的样子
44         i--;
45     }
46     return ans;
47 }
48
49 int main() {
50     int T,n,a[MX];
51     long long ans;
52     scanf("%d",&T);
53     while(T--) {
54         ans=0;
55         root.next[0]=NULL;
56         root.next[1]=NULL;
57         scanf("%d",&n);
58         for(int i=0; i<n; i++) {
59             scanf("%d",&a[i]);
60             BuildTrie(a[i]);
61         }
62         //cout<<"YES1\n";
63         for(int i=0; i<n; i++) {
64             ans=max(ans,Query(a[i]));
65         }
66     //    cout<<"YES2\n";
67         printf("%I64d\n",ans);
68     }
69     return 0;
70 }