CF587F Duff is Mad(AC自动机+根号分治)
CF587F Duff is Mad(AC自动机+根号分治)
题目大意
给定 \(n\) 个字符串 \(s_{1 \dots n}\)。
- \(q\) 次询问 \(s_{l \dots r}\) 在 \(s_k\) 中出现了多少次。
- \(n,q,\sum_{i=1}^n |s_i| \le 10^5\)。
解题思路
相似的题\(s_k\) 在 \(s_{l \dots r}\) 中出现了多少次和此题其他限制条件 AC自动机的题怎么都一个套路啊
很容易想到暴力思路, 离线后直接暴力在原trie树上走, 拿树状数组维护一下就行
很不巧的是\(\sum|s_k|\)并没有保证, 因为一个k可以被询问多次, 因此我们考虑其他做法
然后我们想如果\(|s_k| \le \sqrt m\) m是字符串大小之和, 直接暴力做是可行的, 这样复杂度为\(\Theta(qlog_q+n\sqrt m log_m)\), 可以改变阈值获得更优的复杂度, 但这样也可过
如果\(|s_k| \ge \sqrt m\), 代表着这样的串只有\(\sqrt m\)个, 即使对每个串\(\Theta(n)\)暴力也可以了, 所以我们将询问挂在\(s_k\)上, 将它的每个前缀节点标记为1, dfs一遍求子树权值和, 从一到n加入字符串, 直接加上val[end[i]]就是字符串i的贡献
我的dfs直接写的暴力跳fail, 结果也过了?!
代码如下:
#include <queue>
#include <vector>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MP make_pair
#define ll long long
#define fi first
#define se second
using namespace std;
template <typename T>
void read(T &x) {
x = 0; bool f = 0;
char c = getchar();
for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
for (;isdigit(c);c=getchar()) x=x*10+(c^48);
if (f) x=-x;
}
template <typename T>
inline void Mx(T &x, T y) { x < y && (x = y); }
template <typename T>
inline void Mn(T &x, T y) { x > y && (x = y); }
const int N = 100500;
vector<char> s[N];
char res[N];
int ch[N][26], en[N], dfn[N];
int f[N], cnt, num, T, m, n;
int ne[N], to[N], h[N], tot;
inline void add_e(int x, int y) {
ne[++tot] = h[x], to[h[x] = tot] = y;
}
int add(char *s, int len) {
int p = 0;
for (int i = 1;i <= len; i++) {
int di = s[i] - 'a';
if (!ch[p][di]) ch[p][di] = ++cnt;
p = ch[p][di];
} return p;
}
void build(void) {
queue<int> q;
for (int i = 0;i < 26; i++)
if (ch[0][i]) q.push(ch[0][i]);
while (q.size()) {
int x = q.front(); q.pop();
add_e(f[x], x);
for (int i = 0;i < 26; i++)
if (ch[x][i]) f[ch[x][i]] = ch[f[x]][i], q.push(ch[x][i]);
else ch[x][i] = ch[f[x]][i];
}
}
int siz[N];
void dfs(int x) {
dfn[x] = ++num, siz[x] = 1;
for (int i = h[x]; i; i = ne[i])
dfs(to[i]), siz[x] += siz[to[i]];
}
struct node {
int k, l, num;
bool operator < (const node &i) const {
return l < i.l;
}
};
vector<node> v[N], q;
ll ans[N], d[N], val[N];
void Add(int x, int k) {
for (; x <= num; x += x & -x) d[x] += k;
}
ll Sum(int x) {
ll res = 0;
for (; x ; x -= x & -x) res += d[x];
return res;
}
int main() {
read(n), read(m); T = 250;
for (int i = 1;i <= n; i++) {
scanf ("%s", res + 1);
int t = strlen(res + 1); en[i] = add(res, t);
s[i].push_back('f');
for (int j = 1;j <= t; j++) s[i].push_back(res[j]);
}
build(); dfs(0);
for (int i = 1;i <= m; i++) {
int l, r, k; read(l), read(r), read(k);
if ((int)s[k].size() > T) {
if (l-1) v[k].push_back((node){-1, l-1, i});
v[k].push_back((node){1, r, i});
}
else {
if (l-1) q.push_back((node){-k, l-1, i});
q.push_back((node){k, r, i});
}
}
sort(q.begin(), q.end());
for (int i = 1;i <= n; i++) {
if (!v[i].size()) continue;
sort(v[i].begin(), v[i].end());
memset(val, 0, sizeof(val));
int p = 0;
for (int j = 1;j < (int)s[i].size(); j++) {
p = ch[p][s[i][j]-'a'];
int x = p;
while (x) val[x]++, x = f[x];
}
ll sum = 0, now = 1;
for (auto j: v[i]) {
while (now <= j.l)
sum += val[en[now++]];
ans[j.num] += j.k * sum;
}
}
int now = 1;
for (auto i: q) {
while (now <= i.l) {
int x = en[now];
Add(dfn[x] + siz[x], -1);
Add(dfn[x], 1); now++;
}
int p = 0, f = i.k > 0 ? 1 : -1;
if (i.k < 0) i.k = -i.k;
ll sum = 0;
for (int j = 1;j < (int)s[i.k].size(); j++) {
p = ch[p][s[i.k][j]-'a'];
sum += Sum(dfn[p]);
}
ans[i.num] += f * sum;
}
for (int i = 1;i <= m; i++) printf ("%lld\n", ans[i]);
return 0;
}
