HDU 5536 Chip Factory (枚举+01Trie)

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3143    Accepted Submission(s): 1352

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

231 2 33100 200 300

 

Sample Output

6400

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

Recommend

hujie

 

Statistic | Submit | Discuss | Note

题意：给你n个数(n<=1e3), 求max( (a[i] + a[j]) ^ a[k] ) (i, j, k都不相同)

思路：先将n个数建立01Trie，然后枚举i，j，从01Trie中删除a[i]和a[j]，然后去查询与（a[i]+a[j]）异或的最大值，查询完后再插回去。

另外,删除操作是这样实现的，我们每个节点记录一个cnt值

插入时对所有经过节点的val值加1，删除就将对应节点的cnt值减1。

在树上贪心匹配的时候就只走那些cnt值为正的节点。

#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <algorithm>using namespace std;const int maxn = 1e5 + 5;int a[maxn], ch[maxn][2], cnt[maxn], id, n;char str[15];void Insert(int x){    int rt = 0;    for(int i = 31; i >= 0; i--)    {        int tmp = (x>>i)&1;        if(!ch[rt][tmp])        {            memset(ch[id], 0, sizeof(ch[id]));            cnt[id] = 0;            ch[rt][tmp] = id++;        }        rt = ch[rt][tmp];        cnt[rt]++;    }}void Delete(int x){    int rt = 0;    for(int i = 31; i >= 0; i--)    {        int tmp = (x>>i)&1;        rt = ch[rt][tmp];        cnt[rt]--;    }}int match(int x){    int rt = 0, ans = 0;    for(int i = 31; i >= 0; i--)    {        int tmp = (x>>i)&1;        if(ch[rt][tmp^1] && cnt[ch[rt][tmp^1]])        {            ans |= (1<<i);            rt = ch[rt][tmp^1];        }        else            rt = ch[rt][tmp];    }    return ans;}int main(){    int t;    cin >> t;    while(t--)    {        memset(ch[0], 0, sizeof(ch[0]));        id = 1;        scanf("%d", &n);        for(int i = 1; i <= n; i++)            scanf("%d", &a[i]), Insert(a[i]);        int ans = -2e9;        for(int i = 1; i <= n; i++)        {            Delete(a[i]);            for(int j = i+1; j <= n; j++)            {                Delete(a[j]);                ans = max(ans, match(a[i]+a[j]));                Insert(a[j]);            }            Insert(a[i]);        }        printf("%d\n", ans);    }    return 0;}