hdoj 5536 Chip Factory 【字典树】

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 96    Accepted Submission(s): 51

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

 

题意：给定n个元素，让你从中任意选择三个元素a[i] a[j] a[k] 使得 (a[i] + a[j]) ^ a[k]最大。

思路：建立字典树，枚举a[i]和a[j]，删除后，在Trie上查询a[i]+a[j]能够异或得到的最大值，然后恢复Trie，继续下次查询。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <map>#define MAXN 30100#define LL long long#define Ri(a) scanf("%d", &a)#define Pi(a) printf("%d\n", (a))#define Rl(a) scanf("%lld", &a)#define Pl(a) printf("%lld\n", (a))#define Rs(a) scanf("%s", a)#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define INF 0x3f3f3f3fusing namespace std;struct Tree{    int next[MAXN][2], word[MAXN];    int L, root;    int newnode()    {        for(int i = 0; i < 2; i++)            next[L][i] = -1;        word[L++] = 0;        return L-1;    }    void init()    {        L = 0;        root = newnode();    }    void Insert(int val)    {        int u = root, v;        for(int i = 30; i >= 0; i--)        {            v = (val & (1 << i)) ? 1 : 0;            if(next[u][v] == -1)                next[u][v] = newnode();            u = next[u][v];            word[u]++;        }    }    void Delete(int val)    {        int u = root, v;        for(int i = 30; i >= 0; i--)        {            v = (val & (1 << i)) ? 1 : 0;            u = next[u][v];            word[u]--;        }    }    int Query(int val)    {        int u = root, v;        for(int i = 30; i >= 0; i--)        {            v = (val & (1 << i)) ? 1 : 0;            if(v == 1)            {                if(next[u][0] != -1 && word[next[u][0]])                    u = next[u][0];                else                    u = next[u][1], val ^= (1 << i);            }            else            {                if(next[u][1] != -1 && word[next[u][1]])                    u = next[u][1], val ^= (1 << i);                else                    u = next[u][0];            }        }        return val;    }};Tree tree;int a[1010];int main(){    int t; Ri(t);    W(t)    {        int n; Ri(n);        tree.init();        for(int i = 1; i <= n; i++)            scanf("%d", &a[i]), tree.Insert(a[i]);        int ans = 0;        for(int i = 1; i <= n; i++)        {            tree.Delete(a[i]);            for(int j = i+1; j <= n; j++)            {                tree.Delete(a[j]);                ans = max(ans, tree.Query(a[i]+a[j]));                tree.Insert(a[j]);            }            tree.Insert(a[i]);        }        Pi(ans);    }    return 0;}