Trie

数据结构题单 208. 实现 Trie (前缀树) 总共有 $3 \cdot 10^4$ 次插入，每个单词长度 $2000$，字母个数最多是 $6 \cdot 10^7$ 个，考虑到前缀重复的情况，估算字典树中的节点是 $3 \cdot 10^5$ 个 const int N = 300100; struct TrieTree { int n, idx; vector<VI> son; VI cnt; TrieTree() {} TrieTree(int _n) : n(_n), idx(0) { son = vector<VI>(n, VI(26, -1)); cnt = VI(n, 0); } void insert(string &s) { int p = 0; for (auto c : s) { int u = c - 'a'; if (son[p][u] == -1) son[p][u] = ++idx; p = son[p][u]; } cnt[p]++; } bool search(string &s) { int p = 0; for (auto c : s) { int u = c - 'a'; if (son[p][u] == -1) return false; p = son[p][u]; } return cnt[p] > 0; } bool starts_with(string &s) { int p = 0; for (auto c : s) { int u = c - 'a'; if (son[p][u] == -1) return false; p = son[p][u]; } return true; } }; class Trie { public: TrieTree tr; Trie() { tr = TrieTree(N); } void insert(string word) { tr.insert(word); } bool search(string word) { return tr.search(word); } bool startsWith(string prefix) { return tr.starts_with(prefix); } }; 更好的做法是使用动态创建节点的方式，避免估算不准确。 ...

最长公共后缀查询 要求最长的公共后缀字符串，可以把每个字符串逆序，存到字典树里面，在字典树的节点上保存到达当前节点并且原字符串长度最短的字符串长度和下标。对于给定字符串，直接从字典树中查找当前字符串，把树上能够找到的最后一个节点上的信息返回即可。 初始化的时候，先找到字典树使用到的节点的最大值，只初始化使用到的节点，如果 memset 整个数组会超时。 // Date: Sun Mar 24 22:48:03 2024 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; typedef pair<ll, ll> PLL; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7, MOD1 = 998'244'353; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif // For LeetCode #define LN ListNode #define LNP ListNode * #define TN TreeNode #define TNP TreeNode * #ifdef _DEBUG struct ListNode { int val; ListNode *next; ListNode() : val(0), next(nullptr) {} ListNode(int val) : val(val), next(nullptr) {} ListNode(int val, ListNode *next) : val(val), next(next) {} }; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode() : val(0), left(nullptr), right(nullptr) {} TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} }; #endif // End of LeetCode const int N = 500100; int idx, son[N][26]; struct po { int len, i; }; po val[N]; void Insert(string s, int i) { int p = 0, cur = SZ(s); auto &t = val[p]; if (!t.len || ckmin(t.len, cur)) { t = po{cur, i}; } for (auto c : s) { int u = c - 'a'; if (!son[p][u]) son[p][u] = ++idx; p = son[p][u]; auto &t = val[p]; if (!t.len || ckmin(t.len, cur)) { t = po{cur, i}; } } } int Find(string s) { int p = 0; for (auto c : s) { int u = c - 'a'; if (!son[p][u]) { break; } p = son[p][u]; } return val[p].i; } class Solution { public: vector<int> stringIndices(vector<string> &a, vector<string> &b) { int sum{}; for (auto s : a) sum += SZ(s); idx = 0; For1(i, 0, sum) { For(j, 0, 26) son[i][j] = 0; val[i] = po{0, 0}; } int n = SZ(a); VI res; For(i, 0, n) { string s = a[i]; reverse(all(s)); Insert(s, i); } for (auto s : b) { reverse(all(s)); res.pb(Find(s)); } return res; } }; #ifdef _DEBUG int main(void) { std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); Solution a; return 0; } #endif