HDU 5536 Chip Factory(01字典树找最大异或值)——2015ACM/ICPC亚洲区长春站

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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
   
   
   
   

 

/*********************************************************************/

有一个长度为 n(n≤1000) 的序列 s ，对于不同的 i,j,k ，求

maxi,j,k(si+sj)⊕sk

其中 ⊕ 代表异或运算。

解题思路：据说现场赛的时候 O(n3)是可以过的，然而现场赛重现的时候会送你一个可爱的TLE

此题的正规解法是建一棵字典树，就是Trie树

对 s 构一个Trie树，然后枚举 i,j ，每次删除 si,sj 再查询，再插入回来就可以了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAXN=100010;
int T_T,n,a[MAXN];
struct Trie
{
    int ch[2],size;
}T[MAXN];
int root=1,tot=1;
void Insert(int x)
{
    int o=root;
    T[o].size++;
    for(int k=30;k>=0;k--)
    {
        int c;
        if(x&(1<<k)) c=1;
        else c=0;
        if(!T[o].ch[c]) T[o].ch[c]=++tot;
        o=T[o].ch[c];
        T[o].size++;
    }
}
void Delete(int x)
{
    int o=root;
    T[o].size--;
    for(int k=30;k>=0;k--)
    {
        int c;
        if(x&(1<<k)) c=1;
        else c=0;
        o=T[o].ch[c];
        T[o].size--;
    }
}
int Query(int x)
{
    int o=root;
    for(int k=30;k>=0;k--)
    {
        int c;
        if(x&(1<<k)) c=1;
        else c=0;
        if(c==1)
        {
            if(T[o].ch[0]&&T[T[o].ch[0]].size) o=T[o].ch[0];
            else o=T[o].ch[1],x^=(1<<k);
        }
        else
        {
            if(T[o].ch[1]&&T[T[o].ch[1]].size) o=T[o].ch[1],x^=(1<<k);
            else o=T[o].ch[0];
        }
    }
    return x;
}
int main()
{
    scanf("%d",&T_T);
    while(T_T--)
    {
        int ans=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        for(int i=1;i<=n;i++) Insert(a[i]);
        for(int i=1;i<=n;i++)
        {
            Delete(a[i]);
            for(int j=i+1;j<=n;j++)
            {
                Delete(a[j]);
                ans=max(ans,Query(a[i]+a[j]));
                Insert(a[j]);
            }
            Insert(a[i]);
        }
        printf("%d\n",ans);
        for(int i=1;i<=tot;i++) T[i].ch[0]=0,T[i].ch[1]=0,T[i].size=0;
        tot=1;
    }
    return 0;
}

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