HDU 5536 Chip Factory (暴力 或者 01Trie)

Chip Factory

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

Total Submission(s): 236    Accepted Submission(s): 129

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亚洲区长春站-重现赛（感谢东北师大）

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5536

题目大意：求max( (a[i] + a[j]) ^ a[k] ) (i, j, k都不相同)

题目分析：TMD，长春因为这题错失铜，错失冲银的机会(真是水)

3方暴力可搞，01Trie可搞，爆搜剪枝可搞，结果三小时搞不出来这么一道破题

对面两个女队都过了，Fu*k

暴力 (6s +)

#include <cstdio>
#include <algorithm>
using namespace std;
int a[1005];

int main()
{
	int T;
	scanf("%d", &T);
	while(T --)
	{
		int n, ma = 0;
		scanf("%d" ,&n);
		for(int i = 0; i < n; i++)
			scanf("%d", &a[i]);
		for(int i = 0; i < n; i++)
			for(int j = i + 1; j < n; j++)
				for(int k = j + 1; k < n; k++)
				{
					ma = max(ma, (a[i] + a[j]) ^ a[k]);
					ma = max(ma, (a[i] + a[k]) ^ a[j]);
					ma = max(ma, (a[j] + a[k]) ^ a[i]);
				}
		printf("%d\n", ma);
	}
}

01Tire的时候要注意下标不同，但是数字可能相同

01Trie (3s +)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 1e5;
int a[MAX];

struct Trie
{
	int root, tot, next[MAX][2], cnt[MAX], end[MAX];

	inline int Newnode()
	{
		memset(next[tot], -1, sizeof(next[tot]));
		cnt[tot] = 0;
		end[tot] = 0;
		return tot ++;
	}

	inline void Init()
	{
		tot = 0;
		root = Newnode();
	}

	inline void Insert(int x)
	{
		int p = root;
		cnt[p] ++;
		for(int i = 31; i >= 0; i --)
		{
			int idx = ((1 << i) & x) ? 1 : 0;
			if(next[p][idx] == -1)
				next[p][idx] = Newnode();
			p = next[p][idx];
			cnt[p] ++;
		}
		end[p] = x;
	}

	inline void Del(int x)
	{
		int p = root;
		cnt[p] --;
		for(int i = 31; i >= 0; i --)
		{
			int idx = ((1 << i) & x) ? 1 : 0;
			p = next[p][idx];
			cnt[p] --;
		}
	}

	inline int Search(int x)
	{
		int p = root;
		for(int i = 31; i >= 0; i --)
		{
			int idx = ((1 << i) & x) ? 1 : 0;
			if(idx == 0)
			{
				if(next[p][1] != -1 && cnt[next[p][1]])
					p = next[p][1]; 
				else
					p = next[p][0];
			}
			else
			{
				if(next[p][0] != -1 && cnt[next[p][0]])
					p = next[p][0];
				else
					p = next[p][1];
			}
		}
		return x ^ end[p];
	}
}tr;

int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		tr.Init();
		int n, ma = 0;
		scanf("%d", &n);
		for(int i = 0; i < n; i++)
		{
			scanf("%d", &a[i]);
			tr.Insert(a[i]);
		}
		for(int i = 0; i < n; i++)
		{
			for(int j = i + 1; j < n; j++)
			{
				tr.Del(a[i]);
				tr.Del(a[j]);
				ma = max(ma, tr.Search(a[i] + a[j]));
				tr.Insert(a[i]);
				tr.Insert(a[j]);
			}
		}
		printf("%d\n", ma);
	}
}