[Algorithm basis] Trie tree

1. Trie tree

Trie tree is a data structure for efficiently storing and finding collections of strings.

Two, Trie string statistics

Maintain a collection of strings and support two operations:

1. I x inserts a string x into the collection;

2. Q x asks how many times a string occurs in the set.

There are N � operations in total, and the total length of all input strings does not exceed 10 5 , and the strings only contain lowercase English letters.

input format

The first line contains the integer N , denoting the operand.

Next N lines, each line contains an operation instruction, and the instruction is one of I x or Q x .

output format

For each query instruction Q x , an integer must be output as a result, indicating the number of times x appears in the set.

Each result occupies one line.

data range

1≤N≤2∗104

Input sample:

5
I abc
Q abc
Q ab
I ab
Q ab

Sample output:

1
0
1

code:

#include <iostream>

using namespace std;

const int N = 100010;

// 每个结点的子节点最多为26
//cnt存储以当前单词结尾的单词有多少个
//idx存储当前用到了哪个下标
//下标是0的点即使根节点，又是空结点
int son[N][26], cnt[N], idx; 
char str[N];


void insert(char str[])
{
    int p = 0;
    for(int i = 0;str[i]; i++)
    {
        int u = str[i] - 'a'; //当前字母的编号
        if(!son[p][u]) son[p][u] = ++idx;
        p = son[p][u];
    }
    
    cnt[p]++;
    
}


int query(char str[])
{
    int p = 0;
    for (int i = 0; str[i]; i ++ )
    {
        int u = str[i] - 'a';
        if (!son[p][u]) return 0;
        p = son[p][u];
    }
    return cnt[p];
}

int main()
{
    int n;
    cin>>n;
    while (n -- )
    {
        char op[2];
        cin >> op >> str;
        if(op[0] =='I') insert(str);
        else cout << query(str) <<endl;
    }
    
    return 0;
}

Three, the largest XOR pair

Select two of the given N integers A 1 , A 2 ... A N for xor (exclusive OR) operation, what is the maximum result obtained?

input format

The first line enters an integer N.

The second line inputs N integers A 1 to A N .

output format

Output an integer representing the answer.

data range

1≤N≤105

0≤Ai<231

Input sample:

3
1 2 3

Sample output:

3

code:

1. Violent solution

#include <iostream>
#include <algorithm>

using namespace std;

const int N =1e5;

int n;
int a[N];

int main()
{

  cin >> n;
  for(int i = 0; i < n; i++) cin >> a[i];

  int res = 0;

  for(int i = 0; i < n;i++)
       for (int j = 0; j < n; j ++ )
           res = max(res,a[i]^a[j]);
         
  cout << res << endl;

  return 0;    
}

2.Trie tree solution

#include <iostream>
#include <algorithm>

using namespace std;

const int N =100010, M = 3100010;

int n;

int son[M][2], idx;
int a[N];

void insert(int x)
{
    int p = 0;
    for( int i = 30; i>=0; i--)
    {
        int &s = son[p][x >> i & 1];
        if(!s) s = ++ idx;//创建新结点
        p = s;
    }
}

int query(int x)
{
    int res= 0,p = 0;
    for (int i = 30; i >=0; i-- )
    {
        int s= x>>i &1;
        if(son[p][!s])
        {
            res +=1<<i;
            p = son[p][!s];
            
        }
        else p = son[p][s];
    }
    return res;
}
int main()
{

  cin >> n;
  for(int i = 0; i < n; i++) 
  {
      cin >> a[i];
      insert(a[i]);
  }

  int res = 0;

  for(int i = 0; i < n;i++) res = max(res,query(a[i]));
  cout << res<<endl;
  return 0;    
}