HDU5536：Chip Factory（字典樹）

Chip Factory

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

Total Submission(s): 2699    Accepted Submission(s): 1197

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亞洲區長春站-重制賽（感謝東北師大）

題意：給N個數，找出(a[i]+a[j])^a[k]的最大值，其中i,j,k，為不同的數。

思路：範圍隻用1000，可以枚舉i,j，每次删掉a[i]，a[j]後跑一遍字典樹更新最大值即可。

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
using namespace std;

const int maxn = 1e3+3;
int arr[maxn], cnt;
struct node
{
    int a[2], sum;
}tri[maxn*100];


void ins(int x)
{
    int p = 0;
    for(int i=31; i>=0; --i)
    {
        int t = (x>>i)&1;
        if(!tri[p].a[t])
            tri[p].a[t] = ++cnt;
        p = tri[p].a[t];
        ++tri[p].sum;
    }
}

void del(int x)
{
    int p = 0;
    for(int i=31; i>=0; --i)
    {
        int t = (x>>i)&1;
        p = tri[p].a[t];
        --tri[p].sum;
    }
}

int query(int x)
{
    int p = 0, ans = 0;
    for(int i=31; i>=0; --i)
    {
        int t = (x>>i)&1, rev = tri[p].a[!t], cur = tri[p].a[t];
        if(rev && tri[rev].sum > 0)
            p = rev, ans |= (1<<i);
        else
            p = cur;
    }
    return ans;
}
int main()
{
    int t, n;
    scanf("%d",&t);
    while(t--)
    {
        cnt = 0;
        memset(tri, 0, sizeof(tri));
        scanf("%d",&n);
        for(int i=0; i<n; ++i)
            scanf("%d",&arr[i]), ins(arr[i]);
        int ans = 0;
        for(int i=0; i<n; ++i)
        {
            del(arr[i]);
            for(int j=i+1; j<n; ++j)
            {
                del(arr[j]);
                ans = max(ans ,query(arr[i]+arr[j]));
                ins(arr[j]);
            }
            ins(arr[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}