Trie树总结
理解
试想一种情形,给出一堆字符串,将其根据前缀是否相同而将其组成一颗树,维护两个条件,一是对这颗树的所有结点进行标记,二是在每个字符串结束时做标记.
数组写法
struct Trie
{
int ch[ maxn ][ 26 ]; //结点
int val[ maxn ];
int tot; // 结点标记
void Init()
{
memset( ch, 0, sizeof( ch ) );
memset( val, 0, sizeof( val ) );
tot = 1;
}
void Insert( string s )
{
int root = 0;
for( auto x: s )
{
int Index = x - 'a';
if( !ch[ root ][ Index ] ) ch[ root ][ Index ] = tot ++;
root = ch[ root ][ Index ];
}
val[ root ] = 1;
}
int Search( string s )
{
int root = 0;
for( auto x: s )
{
int Index = x - 'a';
if( !ch[ root ][ Index ] ) return 0;
root = ch[ root ][ Index ];
}
return val[ root ];
}
};
数组写法特点
1.需要分析maxn的取值，最坏情况就是全部散开，即 n * maxwordslength
2.数组写法不适合leetcode
题目列表
- 557E Ann and Half-Palindrome
- 633C Spy Syndrome 2
- 665E Beautiful Subarrays
557E Ann and Half-Palindrome
题意
定义了半回文串，即奇数位置上的数是回文的,给出字符串，求其第k大的半回文子串
题解
首先求出子串的半回文串条件
dp[ i ][ j ] 表示s[ i – j ]中是半回文子串的个数
o(n^2)可以求出.在求出过程中不能直接把半回文子串插入，否则超时.
然后把s的全部子串都插入到Trie中.在Trie中记录当前结点有多少个半回文子串经过
然后dfs求出满足条件的子串
在dfs时，由父串至子串过程中，减去满足在父串停留的满足条件的字符串个数.
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 5010;
const int Max = 2e7;
int dp[ maxn ][ maxn ];
string s;
int k, n;
struct Trie
{
int ch[ Max ][ 2 ];
int num[ Max ];
int tot;
void Init()
{
memset( ch, 0, sizeof( ch ) );
memset( num, 0, sizeof( num ) );
tot = 1;
}
void Insert( int k )
{
int root = 0;
for( int i = k; i < n; ++ i )
{
int Index = s[ i ] - 'a';
if( !ch[ root ][ Index ] ) ch[ root ][ Index ] = tot ++;
root = ch[ root ][ Index ];
if( k == i ) num[ root ] += dp[ k ][ n - 1 ];
else num[ root ] += ( dp[ k ][ n - 1 ] - dp[ k ][ i - 1 ] );
}
}
void Dfs( int root, int k )
{
if( k <= 0 ) return;
int num_a = ch[ root ][ 0 ];
int num_b = ch[ root ][ 1 ];
if( num_a && num[ num_a ] >= k )
{
cout << "a";
if( ch[ num_a ][ 0 ] ) num[ num_a ] -= num[ ch[ num_a ][ 0 ] ];
if( ch[ num_a ][ 1 ] ) num[ num_a ] -= num[ ch[ num_a ][ 1 ] ];
Dfs( num_a, k - num[ num_a ] );
}
else
{
if( num_a ) k -= num[ num_a ];
cout << "b";
if( ch[ num_b ][ 0 ] ) num[ num_b ] -= num[ ch[ num_b ][ 0 ] ];
if( ch[ num_b ][ 1 ] ) num[ num_b ] -= num[ ch[ num_b ][ 1 ] ];
Dfs( num_b, k - num[ num_b ] );
}
}
};
Trie T;
int main()
{
cin >> s;
cin >> k;
n = s.length();
for( int i = 0; i < n; ++ i ) dp[ i ][ i ] = 1;
for( int i = 0; i < n - 1; ++ i ) dp[ i ][ i + 1 ] = ( s[ i ] == s[ i + 1 ] );
for( int i = 2; i < n; ++ i )
for( int j = 0; j + i < n; ++ j )
{
if( s[ j ] == s[ j + i ] ) dp[ j ][ j + i ] = ( j + 2 >= j + i - 2 ? 1 : dp[ j + 2 ][ j + i - 2 ] );
}
for( int i = 1; i < n; ++ i )
for( int j = 0; j < n - 1; ++ j ) dp[ j ][ j + i ] += dp[ j ][ j + i - 1 ];
T.Init();
for( int i = 0; i < n; ++ i ) T.Insert( i );
T.Dfs( 0, k );
return 0;
}
633C Spy Syndrome 2
题意
一堆单词，把这一堆单词都翻转了，拼成了一个串。
然后现在给你一个串，让你找到原来拼的那些单词是什么。
题解
先把当前串s翻转,然后将所有串插入Trie中
数组vis[ i ]标记标记当前s[i]是否到达,然后记录前缀.因为任意一个满足条件的均可以. 所以从前往后更新vis即可.
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
const int Max = 1e5 + 10;
int pre[ Max ], Vis[ Max ], Mappre[ Max ];
map< int, string > Map;
vector< string > goal;
string s, tmp;
int n, m;
struct Trie
{
int ch[ maxn ][ 26 ];
int val[ maxn ];
int tot;
void Init()
{
memset( ch, 0, sizeof( ch ) );
memset( val, 0, sizeof( val ) );
tot = 1;
}
void Insert( string word, int Val )
{
int root = 0;
for( auto x: word )
{
int Index = 0;
if( x >= 'a' && x <= 'z' ) Index = x - 'a';
else Index = x - 'A';
if( !ch[ root ][ Index ] ) ch[ root ][ Index ] = tot ++;
root = ch[ root ][ Index ];
}
val[ root ] = Val;
}
void Search( int now )
{
int root = 0;
for( int i = now; i < n; ++ i )
{
int Index = s[ i ] - 'a';
if( !ch[ root ][ Index ] ) return;
else
{
root = ch[ root ][ Index ];
if( val[ root ] )
{
Vis[ i ] = 1;
pre[ i ] = now;
Mappre[ i ] = val[ root ];
}
}
}
}
};
Trie T;
int main()
{
cin >> n;
cin >> s;
cin >> m;
T.Init();
for( int i = 1; i <= m; ++ i )
{
cin >> tmp;
Map[ i ] = tmp;
T.Insert( tmp, i );
}
reverse( s.begin(), s.end() );
T.Search( 0 );
for( int i = 0; i < n; ++ i ) if( Vis[ i ] ) T.Search( i + 1 );
int num = n - 1;
while( num != -1 )
{
goal.push_back( Map[ Mappre[ num ] ] );
num = pre[ num ] - 1;
}
for( auto x: goal ) cout << x << " ";
return 0;
}
665E Beautiful Subarrays
题意
问你有多少个区间，异或起来大于等于k
题解
pre[x]表示0~x的前缀和
将题意转换为有多少对( x, y )满足 0<= x < y < n， 使得 pre[ x ] ^pre[y] >= k
Trie存储从高位到地位的01,每次将pre[x]放入后,在Trie中遍历pre[ x + 1 ],遇到k的某个二进制为0时，添加和即可.
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e7 + 10;
struct Trie
{
int ch[ maxn ][ 2 ];
int tot_num[ maxn ];
int sz;
void Init()
{
memset( ch, 0, sizeof( ch ) );
memset( tot_num, 0, sizeof( tot_num ) );
sz = 2;
}
void Insert( int num )
{
int cnt = 1;
for( int i = 30; i >= 0; -- i )
{
int Index = ( num >> i ) & 1;
if( !ch[ cnt ][ Index ] ) ch[ cnt ][ Index ] = sz ++;
tot_num[ cnt ] ++;
cnt = ch[ cnt ][ Index ];
}
tot_num[ cnt ] ++;
}
long long Get( int num, int k )
{
int cnt = 1;
long long sum = 0;
for( int i = 30; i >= 0; -- i )
{
int Index_num = ( num >> i ) & 1 ^ 1;
int Index_k = ( k >> i ) & 1;
if( Index_k == 0 )
{
sum +=tot_num[ ch[ cnt ][ Index_num ] ];
cnt = ch[ cnt ][ Index_num ^ 1 ];
}
else cnt = ch[ cnt ][ Index_num ];
}
return sum + tot_num[ cnt ];
}
};
Trie T;
int main()
{
int n, k, x;
cin >> n >> k;
int pre = 0;
long long ans = 0;
T.Init();
T.Insert( 0 );
for( int i = 0; i < n; ++ i )
{
cin >> x;
pre ^= x;
ans += T.Get( pre, k );
T.Insert( pre );
}
cout << ans << endl;
return 0;
}
