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：

#include"iostream"
#include"algorithm"
#include"cstdio"
#include"cmath"
#include"cstring"
#define MX 1400000
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int tree[MX][2],xx;

void BuildTrie(long long a) {
    int i=31;
    int p=0;
    while(0<=i) {
        bool num=a&(1<<i);
        if(!tree[p][num]) {        //如果节点为空 
            tree[p][num]=++xx;//标记并创建新的子节点 
        }
        p=tree[p][num];
        //    cout<<"YES"<<i<<"   B is:"<<num<<"
"; //建立的Trie树的样子
        i--;
    }
}

long long Query(long long  a) {
    int i=31,p=0;
    long long ans=0;
    while(0<=i) {
        bool num=!(a&(1<<i));  //取反查找
        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
        else p = tree[p][!num];  //按照相同的顺序找
    //    cout<<"YES"<<i<<"   B is:"<<num<<"     ans is:"<<ans<<endl;   //查询时候的Trie树的样子
        i--;
    }
    return ans;
}

int main() {
    int T,n;
    long long a[100050],ans;
    scanf("%d",&T);
    while(T--) {
        ans=0;
        memset(tree,0,sizeof(tree));
        xx=0;
        scanf("%d",&n);
        for(int i=1; i<=n; i++) {
            scanf("%I64d",&a[i]);
            BuildTrie(a[i]);
        }
        //cout<<"YES1
";
        for(int i=1; i<=n; i++) {
            ans=max(ans,Query(a[i]));
        }
        //    cout<<"YES2
";
        printf("%I64d
",ans);
    }
    return 0;
}

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。

#include"iostream"
#include"algorithm"
#include"cstdio"
#include"cmath"
#include"cstring"
#define MX 110000
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;

struct Trie {
    Trie *next[2];
} root;

void BuildTrie(int a) {
    Trie *p=&root,*q;
    int i=31;
    while(i>=0) {
    bool num=a&(1<<i);
        if(p->next[num]==NULL) {
            q=(Trie *)malloc(sizeof(root));//申请一块新内存; //动态分配内存
            q->next[1]=NULL;
            q->next[0]=NULL;    //清空申请内存的所有子节点
            p->next[num]=q;        //往子节点下去继续记录字典树
            p=p->next[num];
        } else {
            p=p->next[num];
        }
    //    cout<<"YES"<<i<<"   B is:"<<num<<"
"; //建立的Trie树的样子 
        i--;
    }
}

long long Query(int a) {
    Trie *p=&root;
    long long ans=0;
    int i=31;
    while(0<=i) {
        bool num=a&(1<<i);  
        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
        else if(p->next[num]!=NULL) p=p->next[num];  //按照相同的顺序找
    //    cout<<"YES"<<i<<"   B is:"<<num<<"
"<<"ans is:"<<ans<<endl;   //查询时候的Trie树的样子 
        i--;
    }
    return ans;
}

int main() {
    int T,n,a[MX];
    long long ans;
    scanf("%d",&T);
    while(T--) {
        ans=0;
        root.next[0]=NULL;
        root.next[1]=NULL;
        scanf("%d",&n);
        for(int i=0; i<n; i++) {
            scanf("%d",&a[i]);
            BuildTrie(a[i]);
        }
        //cout<<"YES1
"; 
        for(int i=0; i<n; i++) {
            ans=max(ans,Query(a[i]));
        }
    //    cout<<"YES2
"; 
        printf("%I64d
",ans);
    }
    return 0;
}