洛谷 P5829 【模板】失配树

洛谷传送门

思路

首先发现，题中所述的 \(\mathrm{border}\) 就是 KMP 中的 fail。求两个前缀的公共 \(\mathrm{border}\)，就是跳到最大的公共的 \(\mathrm{fail}\)。

因此建出 \(\mathrm{fail}\) 树，则每次询问跑一遍 LCA 即可。可以用树剖、倍增或 tarjan 实现。

注意 \(\mathrm{border}\) 不能是自己，因此要先往上跳一步再求 LCA。

代码

code

/*

p_b_p_b txdy
AThousandMoon txdy
AThousandSuns txdy
hxy txdy

*/

#include <bits/stdc++.h>
#define pb push_back
#define fst first
#define scd second

using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;

const int maxn = 1000100;

int n, m, head[maxn], len, fail[maxn];
int sz[maxn], fa[maxn], son[maxn], dp[maxn];
int top[maxn];
bool vis[maxn];
char s[maxn];

struct edge {
	int to, next;
} edges[maxn << 1];

void add_edge(int u, int v) {
	edges[++len].to = v;
	edges[len].next = head[u];
	head[u] = len;
}

int dfs(int u, int f, int d) {
	dp[u] = d;
	fa[u] = f;
	sz[u] = 1;
	int maxson = -1;
	for (int i = head[u]; i; i = edges[i].next) {
		int v = edges[i].to;
		if (v == f) {
			continue;
		}
		sz[u] += dfs(v, u, d + 1);
		if (sz[v] > maxson) {
			son[u] = v;
			maxson = sz[v];
		}
	}
	return sz[u];
}

void dfs2(int u, int tp) {
	top[u] = tp;
	vis[u] = 1;
	if (!son[u]) {
		return;
	}
	dfs2(son[u], tp);
	for (int i = head[u]; i; i = edges[i].next) {
		int v = edges[i].to;
		if (!vis[v]) {
			dfs2(v, v);
		}
	}
}

int querylca(int x, int y) {
	while (top[x] != top[y]) {
		if (dp[top[x]] < dp[top[y]]) {
			swap(x, y);
		}
		x = fa[top[x]];
	}
	if (dp[x] > dp[y]) {
		swap(x, y);
	}
	return x;
}

void solve() {
	scanf("%s%d", s + 1, &m);
	n = strlen(s + 1);
	for (int i = 2, j = 0; i <= n; ++i) {
		while (j && s[i] != s[j + 1]) {
			j = fail[j];
		}
		if (s[i] == s[j + 1]) {
			++j;
		}
		fail[i] = j;
	}
	for (int i = 1; i <= n; ++i) {
		add_edge(fail[i], i);
	}
	dfs(0, -1, 1);
	dfs2(0, 0);
	while (m--) {
		int x, y;
		scanf("%d%d", &x, &y);
		printf("%d\n", querylca(fa[x], fa[y]));
	}
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}