HDU5536 Chip Factory

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 6984    Accepted Submission(s): 3129

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

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

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题解：01字典树.

 

参考代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;    
}
int const maxn=1e5+10;
int a[maxn];
struct Trie_01{
    int root,tot,nxt[maxn][2],cnt[maxn],end[maxn];
    int Newnode()
    {
        memset(nxt[tot],-1,sizeof(nxt[tot]));
        cnt[tot]=0;
        end[tot]=0;
        return tot++;
    }
    void Init()
    {
        tot=0;
        root=Newnode();
    }
    void Insert(int x)
    {
        int p=root;
        cnt[p]++;
        for(int i=31;i>=0;--i)
        {
            int idx=(1&(x>>i));
            if(nxt[p][idx]==-1)
                nxt[p][idx]=Newnode();
            p=nxt[p][idx];
            ++cnt[p];//可能会有重复的 
        }
        end[p]=x;
    }
    void Delete(int x)//删除一个数x 
    {
        int p=root;
        --cnt[p];
        for(int i=31;i>=0;i--)
        {
            int idx=(1&(x>>i));
            p=nxt[p][idx];
            --cnt[p];
        }
    }
    int QueryMax(int x)//求xor最大值 
    {
        int p=root;
        for(int i=31;i>=0;--i)
        {
            int idx=(1&(x>>i));
            if(idx==0)
            {
                if(nxt[p][1]!=-1&&cnt[nxt[p][1]]) p=nxt[p][1]; 
                else p=nxt[p][0];
            }
            else
            {
                if(nxt[p][0]!=-1&&cnt[nxt[p][0]]) p=nxt[p][0];
                else p=nxt[p][1];
            }
        }
        return (x^end[p]);
    }
    int QueryMin(int x)//求xor最小值
    {
        int p=root;    
        for(int i=31;i;--i)
        {
            int idx=(1&(x>>i));
            if(idx==1)
            {
                if(nxt[p][1]!=-1&&cnt[nxt[p][1]]) p=nxt[p][1]; 
                else p=nxt[p][0];
            }
            else
            {
                if(nxt[p][0]!=-1&&cnt[nxt[p][0]]) p=nxt[p][0];
                else p=nxt[p][1];
            }
        }
        return (x^end[p]);
    }
}trie;
 
int main()
{
    int T=read();
    while(T--)
    {
        trie.Init();
        int n,ans=0;
        n=read();
        for(int i=0;i<n;++i)
        {
            a[i]=read();
            trie.Insert(a[i]);
        }
        for(int i=0;i<n;++i)
        {
            for(int j=i+1;j<n;++j)
            {
                trie.Delete(a[i]);
                trie.Delete(a[j]);
                ans=max(ans,trie.QueryMax(a[i]+a[j]));
                trie.Insert(a[i]);
                trie.Insert(a[j]);
            }
        }
        printf("%d
", ans);
    }
    return 0;
}