CF-1207 G.Indie Album(Trie上跑AC自動機)

題目鏈接

題意

$n$個串,串的生成方式有兩種:

單獨一個字符

在上一個串的基礎上加一個字符

$q$個詢問,問第$i$個串中出現串$s$的次數

思路

離線處理詢問,把詢問的字符串建AC自動機,并得到$fail$樹

對于一個字符串$s$出現的次數等于$fail$書上對應的節點和它所有子節點出現次數的和 ($AC$自動機性質)

在$Trie$上$dfs$并把對應在$AC$自動機上的節點加1(樹狀數組),當$dfs$走到字符的終點(當前節點帶有詢問),處理結果.

節點和子樹的和可以用樹狀數組和$dfs$序得到

#include <bits/stdc++.h> using namespace std; const int maxn = 5e5 + 5; int n; char s[maxn]; int ans[maxn]; vector<pair<int,int> > Q[maxn]; struct Bit{int c[maxn];void init() {fill(c, c+maxn, 0);}int lowbit(int x) {return x & (-x);}void add(int x, int d) {while (x < maxn) {c[x] += d;x += lowbit(x);}}int getsum(int x) {int sum = 0;while (x) {sum += c[x];x -= lowbit(x);}return sum;}int query(int l, int r) {return getsum(r) - getsum(l-1);} }bit; struct Trie{int nex[maxn][26], idx[maxn];int root, p;int newnode() {fill(nex[p], nex[p]+26, 0);return p++;}void init() {p = 0;root = newnode();}void add(int id, int u, int c) {u = idx[u];if (nex[u][c-'a'] == 0) nex[u][c-'a'] = newnode();idx[id] = nex[u][c-'a'];} }trie; struct AC{int nex[maxn][26], fail[maxn];int root, p;vector<int> g[maxn];int in[maxn], out[maxn], cnt;int newnode() {for (int i = 0; i < 26; ++i) nex[p][i] = -1;return p++;}void init() {p = 0;cnt = 0;root = newnode();}int insert(char *buf) {int now = root;for (int i = 0; buf[i]; ++i) {if (nex[now][buf[i]-'a'] == -1) {nex[now][buf[i]-'a'] = newnode();}now = nex[now][buf[i]-'a'];}return now;}void build() {queue<int> que;fail[root] = root;for (int i = 0; i < 26; ++i) {if (nex[root][i] == -1) nex[root][i] = root;else {fail[nex[root][i]] == root;que.push(nex[root][i]);}}while (!que.empty()) {int now = que.front();que.pop();g[fail[now]].push_back(now); // 求fail樹for (int i = 0; i < 26; ++i) {if (nex[now][i] == -1) nex[now][i] = nex[fail[now]][i];else {fail[nex[now][i]] = nex[fail[now]][i];que.push(nex[now][i]);}}}}void dfs(int u) { // 求dfs序in[u] = ++cnt;for (int i = 0; i < (int)g[u].size(); ++i) {int v = g[u][i];dfs(v);}out[u] = cnt;}void dfs(int u1, int u2) {bit.add(in[u2], 1); // 對應AC節點++for (int j = 0; j < (int)Q[u1].size(); ++j) { // Trie節點對應有詢問int pos = Q[u1][j].first;int id = Q[u1][j].second;ans[id] = bit.query(in[pos], out[pos]);}for (int i = 0; i < 26; ++i) {if (trie.nex[u1][i] == 0) continue;dfs(trie.nex[u1][i], nex[u2][i]);}bit.add(in[u2], -1); // 去掉之前標記}void solve() {build();dfs(0); // dfs序dfs(0, 0);} }ac; int main() {scanf("%d", &n);trie.init();bit.init();ac.init();for (int i = 1; i <= n; ++i) {int op, num;char c;scanf("%d", &op);if (op == 2) {scanf("%d %c", &num, &c);trie.add(i, num, c);}else {scanf(" %c", &c);trie.add(i, 0, c);}}int q;scanf("%d", &q);for (int i = 1; i <= q; ++i){int id;scanf("%d %s", &id, s);Q[trie.idx[id]].push_back(make_pair(ac.insert(s), i));}ac.solve();for (int i = 1; i <= q; ++i) {printf("%d\n", ans[i]);}return 0; }

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