hdoj 5536 Chip Factory 【字典树】_acm chip factory-优快云博客

本文链接：https://blog.youkuaiyun.com/chenzhenyu123456/article/details/49562945

本文介绍了一种解决芯片工厂中特定问题的方法，该问题要求从一批芯片中选取三个不同的芯片，使得它们之间的特定运算结果达到最大值。通过构建字典树（Trie树）并运用枚举技巧，有效地解决了这一问题。

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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:

max i, j, k (s i + s j) \oplus s k

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 0x3f3f3f3f
using 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;
}