Chip Factory（01字典树）

Chip Factory

http://acm.hdu.edu.cn/showproblem.php?pid=5536

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 280    Accepted Submission(s): 158

Problem Description

John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage large amounts of products, every processor has a serial number. More specifically, the factory produces n chips today, the i-th chip produced this day has a serial number si.

At the end of the day, he packages all the chips produced this day, and send it to wholesalers. More specially, he writes a checksum number on the package, this checksum is defined as below:

maxi,j,k(si+sj)⊕sk

which i,j,k are three different integers between 1 and n. And ⊕ is symbol of bitwise XOR.

Can you help John calculate the checksum number of today?

Input

The first line of input contains an integer T indicating the total number of test cases.

The first line of each test case is an integer n, indicating the number of chips produced today. The next line has n integers s1,s2,..,sn, separated with single space, indicating serial number of each chip.

1≤T≤1000
3≤n≤1000
0≤si≤109
There are at most 10 testcases with n>100

Output

For each test case, please output an integer indicating the checksum number in a line.

Sample Input

2

3

1 2 3

3

100 200 300 

Sample Output

6

400

用num标记该点是否被删除，剩下的就是01字典树的模板了

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<algorithm>
 #include<queue>
 #define maxn 1005
 using namespace std;
 int tol;
 long long val[maxn*];
 int tree[maxn*][];
 int num[maxn*][];
 long long a[maxn];

 void init(){
     tol=;
     tree[][]=tree[][]=;
     memset(num,,sizeof(num));
 }

 void add(long long x,int k){
     int u=;
     for(int i=;i>=;i--){
         int v=(x>>i)&;
         num[u][v]+=k;
         if(!tree[u][v]){
             tree[tol][]=tree[tol][]=;
             val[tol]=-;
             tree[u][v]=tol++;
         }
         u=tree[u][v];
     }
     val[u]=x;
 }

 long long query(long long n){
     int u=;
     for(int i=;i>=;i--){
         int v=(n>>i)&;
         if(num[u][v^]) u=tree[u][v^];
         else u=tree[u][v];
     }
     return val[u];
 }

 int main(){
     int t;
     scanf("%d",&t);
     while(t--){
         int n;
         scanf("%d",&n);
         init();
         for(int i=;i<=n;i++){
             scanf("%lld",&a[i]);
             add(a[i],);
         }
         long long tmp;
         long long ans=;
         for(int i=;i<=n;i++){
             add(a[i],-);
             for(int j=i+;j<=n;j++){
                 add(a[j],-);
                 tmp=a[i]+a[j];
                 ans=max(ans,tmp^query(tmp));
                 add(a[j],);
             }
             add(a[i],);
         }
         printf("%lld\n",ans);
     }

 }