【BZOJ4477】【JSOI2015】字符串树

【题目链接】

• 点击打开链接
  

【思路要点】

• 用可持久化字典树维护每个点到根路径上字符串的字典树。
• 询问时分别查询\(x\)，\(y\)和\(Lca(x,y)\)即可。
• 时间复杂度\(O((N+Q)(|S|+LogN))\)。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
const int MAXLOG = 18;
const int MAXP = 1e6 + 5;
const int MAXL = 15;
const int MAXC = 26;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
struct Trie {
	struct Node {
		int child[MAXC];
		int size;
	} a[MAXP];
	int size;
	int extend(int root, char *s) {
		int now = ++size, ans = size, len = strlen(s + 1);
		a[now] = a[root];
		a[now].size++;
		for (int i = 1; i <= len; i++) {
			int tmp = s[i] - 'a';
			a[now].child[tmp] = ++size;
			root = a[root].child[tmp];
			a[size] = a[root];
			now = a[now].child[tmp];
			a[now].size++;
		}
		return ans;
	}
	int query(int root, char *s) {
		int now = root, len = strlen(s + 1);
		for (int i = 1; i <= len; i++)
			now = a[now].child[s[i] - 'a'];
		return a[now].size;
	}
} Trie;
struct edge {int dest, home; };
int n, m, root[MAXN];
char s[MAXL], st[MAXN][MAXL];
int depth[MAXN], father[MAXN][MAXLOG];
vector <edge> a[MAXN];
void dfs(int pos, int fa, int from) {
	depth[pos] = depth[fa] + 1;
	father[pos][0] = fa;
	if (fa != 0) root[pos] = Trie.extend(root[fa], st[from]);
	for (int i = 1; i < MAXLOG; i++)
		father[pos][i] = father[father[pos][i - 1]][i - 1];
	for (unsigned i = 0; i < a[pos].size(); i++)
		if (a[pos][i].dest != fa) dfs(a[pos][i].dest, pos, a[pos][i].home);
}
int lca(int x, int y) {
	if (depth[x] < depth[y]) swap(x, y);
	for (int i = MAXLOG - 1; i >= 0; i--)
		if (depth[father[x][i]] >= depth[y]) x = father[x][i];
	if (x == y) return x;
	for (int i = MAXLOG - 1; i >= 0; i--)
		if (father[x][i] != father[y][i]) {
			x = father[x][i];
			y = father[y][i];
		}
	return father[x][0];
}
int main() {
	read(n);
	for (int i = 1; i <= n - 1; i++) {
		int x, y; read(x), read(y);
		a[x].push_back((edge) {y, i});
		a[y].push_back((edge) {x, i});
		scanf(" %s", st[i] + 1);
	}
	dfs(1, 0, 0);
	read(m);
	for (int i = 1; i <= m; i++) {
		int x, y; read(x), read(y);
		scanf("%s", s + 1);
		int Lca = lca(x, y);
		writeln(Trie.query(root[x], s) + Trie.query(root[y], s) - 2 * Trie.query(root[Lca], s));
	}
	return 0;
}