HDU5536：Chip Factory（字典树）

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 ⊕<
14210
/span> 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）

题意：给N个数，找出(a[i]+a[j])^a[k]的最大值，其中i,j,k，为不同的数。

思路：范围只用1000，可以枚举i,j，每次删掉a[i]，a[j]后跑一遍字典树更新最大值即可。

# 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;
}