HDU 5536 Chip Factory 【01字典树删除】

题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=5536

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 4915    Accepted Submission(s): 2205

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

 

题目大意：

给 N 个数，在这 N 个数里找到三个值 i， j，k 使得（i+j）⊕ k 最大，输出这个最大值。

解题思路：

每次在01字典树里删除 i 和 j 然后 （i+j）和01字典树里的匹配求最大异或值。

AC code：

#include <bits/stdc++.h>
#define ll long long int
#define INF 0x3f3f3f3f
using namespace std;

const int MAXN = 1e3+10;
int ch[MAXN*32][2];
ll value[MAXN*32];
ll b[MAXN];
int num[MAXN*32];
int node_cnt;

void init()
{
    memset(ch[0], 0, sizeof(ch[0]));
    node_cnt = 1;
}

void Insert(ll x)
{
    int cur = 0;
    for(int i = 32; i >= 0; i--){
        int index = (x>>i)&1;
        if(!ch[cur][index]){
            memset(ch[node_cnt], 0, sizeof(ch[node_cnt]));
            ch[cur][index] = node_cnt;
            value[node_cnt] = 0;
            num[node_cnt++] = 0;
        }
        cur = ch[cur][index];
        num[cur]++;
    }
    value[cur] = x;
}

void update(ll x, int d)
{
    int cur = 0;
    for(int i =32; i >= 0; i--){
        int index = (x>>i)&1;
        cur = ch[cur][index];
        num[cur]+=d;
    }
}

ll query(ll x)
{
    int cur = 0;
    for(int i = 32; i >= 0; i--)
    {
        int index = (x>>i)&1;
        if(ch[cur][index^1] && num[ch[cur][index^1]]) cur = ch[cur][index^1];
        else cur = ch[cur][index];
    }
    return x^value[cur];
}

int main()
{
    int T_case, N;
    scanf("%d", &T_case);
    while(T_case--)
    {
        scanf("%d", &N);
        init();
        for(int i = 1; i <= N; i++)
        {
            scanf("%lld", &b[i]);
            Insert(b[i]);
        }
        ll ans = 0;
        for(int i = 1; i <= N; i++){
            for(int j = 1; j <= N; j++){
                if(i == j) continue;
                update(b[i], -1);
                update(b[j], -1);
                ans = max(ans, query(b[i]+b[j]));
                update(b[i], 1);
                update(b[j], 1);
            }
        }
        printf("%lld
", ans);
    }
    return 0;
}