hdu 5536 Chip Factory (01 Trie)

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5536

题面;

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 6277    Accepted Submission(s): 2847

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

模板题

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

实现代码；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 1e3+;
int tot;
int ch[*M][],vis[*M];
ll val[*M],a[M];

void init(){
    memset(vis,,sizeof(vis));
    tot = ;
    ch[][] = ch[][] = ;
}

void ins(ll x){
    int u = ;
    for(int i = ;i >= ;i --){
        int v = (x>>i)&;
        if(!ch[u][v]){
            ch[tot][] = ch[tot][] = ;
            val[tot] = ;
            vis[tot] = ;
            ch[u][v] = tot++;
        }
        u = ch[u][v];
        vis[u]++;
    }
    val[u] = x;
}

void update(ll x,int c){
    int u = ;
    for(int i = ;i >= ;i --){
        int v = (x>>i)&;
        u = ch[u][v];
        vis[u] += c;
    }
}

ll query(ll x){
    int u = ;
    for(int i = ;i >= ;i --){
        int v = (x>>i)&;
        if(ch[u][v^]&&vis[ch[u][v^]]) u = ch[u][v^];
        else u = ch[u][v];
    }
    return x^val[u];
}

int main()
{
    ios::sync_with_stdio();
    cin.tie(); cout.tie();
    int t,n,m;
    cin>>t;
    while(t--){
        cin>>n;
        ll mx = ;
        init();
        for(int i = ;i <= n;i ++)
            cin>>a[i],ins(a[i]);
        for(int i = ;i <= n;i ++){
            for(int j = i+;j <= n;j ++){
                update(a[i],-); update(a[j],-);
                mx = max(mx,query(a[i]+a[j]));
                update(a[i],); update(a[j],);
            }
        }
        cout<<mx<<endl;
    }
}