CodeForces 1483F Exam

洛谷传送门

CF 传送门

ACAM 好题。

考虑枚举长串，在长串上枚举短串的右端点。显然符合条件的短串为最长的串，也即在 fail 树上往上跳到的第一个为某个串结尾的串。

还要保证这个串不被其他的串包含，简单特判即可。

那么最后所有对答案造成贡献的短串为 算到次数 \(=\) 在长串的出现次数 的所有短串。

出现次数可以用树状数组维护。

感觉思路都很自然啊！

code

// Problem: F. Exam
// Contest: Codeforces - Codeforces Round #709 (Div. 1, based on Technocup 2021 Final Round)
// URL: https://codeforces.com/problemset/problem/1483/F
// Memory Limit: 512 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<int, int> pii;

const int maxn = 1000100;

int n, head[maxn], len;
string s[maxn];

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

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

int ch[maxn][26], fail[maxn], ntot, id[maxn], di[maxn], a[maxn], bl[maxn];

void insert(string s, int k) {
	int p = 0;
	for (char c : s) {
		if (!ch[p][c - 'a']) {
			ch[p][c - 'a'] = ++ntot;
		}
		p = ch[p][c - 'a'];
	}
	id[p] = k;
	di[k] = p;
}

void build() {
	queue<int> q;
	for (int i = 0; i < 26; ++i) {
		if (ch[0][i]) {
			q.push(ch[0][i]);
		}
	}
	while (q.size()) {
		int u = q.front();
		q.pop();
		for (int i = 0; i < 26; ++i) {
			if (ch[u][i]) {
				fail[ch[u][i]] = ch[fail[u]][i];
				q.push(ch[u][i]);
			} else {
				ch[u][i] = ch[fail[u]][i];
			}
		}
	}
	for (int i = 1; i <= ntot; ++i) {
		add_edge(fail[i], i);
	}
}

int st[maxn], times, ed[maxn], up[maxn];

void dfs(int u) {
	if (id[u]) {
		up[u] = u;
	}
	st[u] = ++times;
	for (int i = head[u]; i; i = edges[i].next) {
		int v = edges[i].to;
		up[v] = up[u];
		dfs(v);
	}
	ed[u] = times;
}

int c[maxn];

void update(int x, int d) {
	for (int i = x; i < maxn; i += (i & (-i))) {
		c[i] += d;
	}
}

int query(int x) {
	int res = 0;
	for (int i = x; i; i -= (i & (-i))) {
		res += c[i];
	}
	return res;
}

void solve() {
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		cin >> s[i];
		insert(s[i], i);
	}
	build();
	dfs(0);
	int ans = 0;
	for (int i = 1; i <= n; ++i) {
		int p = 0;
		for (int j = 0; j < (int)s[i].size(); ++j) {
			p = ch[p][s[i][j] - 'a'];
			update(st[p], 1);
			if (j == (int)s[i].size() - 1) {
				p = fail[p];
			}
			if (!id[up[p]]) {
				bl[j] = 2e9;
			} else {
				bl[j] = j - (int)s[id[up[p]]].size() + 1;
				a[j] = id[up[p]];
			}
		}
		int mn = 1e9;
		map<int, int> mp;
		for (int j = (int)s[i].size() - 1; ~j; --j) {
			if (bl[j] < mn) {
				mn = bl[j];
				++mp[a[j]];
			}
		}
		for (pii p : mp) {
			int k = di[p.fst];
			if (query(ed[k]) - query(st[k] - 1) == p.scd) {
				++ans;
			}
		}
		p = 0;
		for (char c : s[i]) {
			p = ch[p][c - 'a'];
			update(st[p], -1);
		}
	}
	printf("%d\n", ans);
}

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