HDU - 5536 Chip Factory 字典树的删除

Posted 2023-02-07

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 6090    Accepted Submission(s): 2745

 

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）​​

 

 

Recommend

hujie

题意：n个数，求任取两个数的加和，与剩下的数异或的最大值

分析：

01字典树求异或的最大值，需要删除操作,因为01字典树，我们可以对每一个节点进行标记，num[u]表示插入字符的01串经过该节点的次数，删除的时候，遍历数字二进制所对应的节点01，使其num【u】-1就可以了。（其实num跟统计前缀差不多作用）

 

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=1005;
const int maxnode=1000000+100;
int num[maxnode];
const int sigma_size=3;
struct Trie

    int ch[maxnode][sigma_size];
    int sz;
    void init()
    
        sz=1;
        memset(ch[0],0,sizeof(ch[0]));
    
    void insert(ll x)
    
        int u=0;
        for(int i=32;i>=0;i--)
        
            int id=(x>>i)&1;//取出x二进制第i位
            if(ch[u][id]==0)
            
                ch[u][id]=sz;
                
                memset(ch[sz],0,sizeof(ch[sz]));
                num[sz]=0;
                sz++;
            
            u=ch[u][id];
            num[u]++;
      
        
        
    
    void update(ll x, int d)
   
     int u = 0;
     for(int i =32; i >= 0; i--)
        int index = (x>>i)&1;
        u = ch[u][index];
        num[u]+=d;
    
    
    ll find(ll x)
    
      ll ans=0;
      int u=0;
        for(int i=32;i>=0;i--)
        
            int id=(x>>i)&1;
            int y=id^1;    //找与id不同的
            if(ch[u][y]&&num[ch[u][y]])
            
          u=ch[u][y];
          ans=ans<<1|1;  //不同为1,ans=ans*2+1
            
            else
         
         u=ch[u][id];
          ans<<=1;      //相同为0,ans=ans*2
            
         
        
        return ans;
    
;
Trie trie;
ll a[N];
int main()

   int t;
   scanf("%d",&t);
   while(t--)
   
    ll  maxx=0;
    int n;
   trie.init();
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
   
      scanf("%lld",&a[i]); 
      trie.insert(a[i]);
   
   for(int i=1;i<n;i++)
      for(int j=i+1;j<=n;j++)
     
      trie.update(a[i],-1);
      trie.update(a[j],-1);
      maxx=max(maxx,trie.find(a[i]+a[j]));
      trie.update(a[i],1);
      trie.update(a[j],1);
        
   
   cout<<maxx<<endl;
   
   
    return 0;