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*35];
int tree[maxn*35][2];
int num[maxn*32][2];
long long a[maxn];

void init(){
    tol=1;
    tree[0][0]=tree[0][1]=0;
    memset(num,0,sizeof(num));
}

void add(long long x,int k){
    int u=0;
    for(int i=32;i>=0;i--){
        int v=(x>>i)&1;
        num[u][v]+=k;
        if(!tree[u][v]){
            tree[tol][0]=tree[tol][1]=0;
            val[tol]=-1;
            tree[u][v]=tol++;
        }
        u=tree[u][v];
    }
    val[u]=x;
}

long long query(long long n){
    int u=0;
    for(int i=32;i>=0;i--){
        int v=(n>>i)&1;
        if(num[u][v^1]) u=tree[u][v^1];
        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=1;i<=n;i++){
            scanf("%lld",&a[i]);
            add(a[i],1);
        }
        long long tmp;
        long long ans=0;
        for(int i=1;i<=n;i++){
            add(a[i],-1);
            for(int j=i+1;j<=n;j++){
                add(a[j],-1);
                tmp=a[i]+a[j];
                ans=max(ans,tmp^query(tmp));
                add(a[j],1);
            }
            add(a[i],1);
        }
        printf("%lld
",ans);
    }

}