hdu5536 Chip Factory xor，字典树

hdu5536 Chip Factory   xor，字典树

Chip Factory

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

 

看到异或就应该想到字典树。。。

注意用siz来处理删除，而不要直接删除结点。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=(1<<29);
const double EPS=0.000000001;
const double Pi=acos(-1.0);

struct Trie
{
    int ch[2];
    int siz;
    void debug()
    {
        printf("lch=%2d rch=%2d siz=%2d\n",ch[1],ch[0],siz);
    }
};Trie tr[maxn];int p;int rt;
int n,a[maxn];

int newnode()
{
    tr[++p].siz=0;
    tr[p].ch[0]=tr[p].ch[1]=-1;
    return p;
}

void Init()
{
    rt=0;
    tr[rt].ch[0]=tr[rt].ch[1]=-1;
    tr[rt].siz=0;
    p=0;
}

void Insert(int t)
{
    int k=rt;
    tr[k].siz++;
    int c;
    for(int i=30;i>=0;i--){
        if(t&(1<<i)) c=1;
        else c=0;
        if(tr[k].ch[c]==-1) tr[k].ch[c]=newnode();
        k=tr[k].ch[c];
        tr[k].siz++;
    }
}

void Delete(int t)
{
    int k=rt;
    tr[k].siz--;
    int c;
    for(int i=30;i>=0;i--){
        if(t&(1<<i)) c=1;
        else c=0;
        k=tr[k].ch[c];
        tr[k].siz--;
    }
}

int query(int t)
{
    int k=rt;
    int res=0;
    for(int i=30;i>=0;i--){
        if(t&(1<<i)){
            if(tr[k].ch[0]==-1||tr[tr[k].ch[0]].siz==0){
                k=tr[k].ch[1];
            }
            else{
                k=tr[k].ch[0];
                res|=(1<<i);
            }
        }
        else{
            if(tr[k].ch[1]==-1||tr[tr[k].ch[1]].siz==0){
                k=tr[k].ch[0];
            }
            else{
                k=tr[k].ch[1];
                res|=(1<<i);
            }
        }
    }
    return res;
}

int main()
{
    freopen("in.txt","r",stdin);
    int T;cin>>T;
    while(T--){
        scanf("%d",&n);
        Init();
        REP(i,1,n) scanf("%d",&a[i]),Insert(a[i]);
        int ans=0;
        REP(i,1,n){
            REP(j,i+1,n){
                Delete(a[i]);Delete(a[j]);
                int tmp=query(a[i]+a[j]);
                ans=max(ans,tmp);
                Insert(a[i]);Insert(a[j]);
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}