BZOJ3413: 匹配(后缀自动机 线段树合并)

时间：2023-02-12 19:56:32

题意

题目链接

Sol

神仙题Orz

后缀自动机 + 线段树合并。。。

首先可以转化一下模型(想不到qwq)：问题可以转化为统计\(B\)中每个前缀在\(A\)中出现的次数。(画一画就出来了)

然后直接对\(A\)串建SAM，线段树合并维护一下siz就行了

#include<bits/stdc++.h>

using namespace std;

const int MAXN = 4e5 + 10, SS = 1e7 + 10;

int N, M;

char S[MAXN], T[MAXN];

int fa[MAXN], len[MAXN], ch[MAXN][11], root = 1, las = 1, tot = 1;

vector<int> par[MAXN];

int insert(int x) {

	int now = ++tot, pre = las; las = now; len[now] = len[pre] + 1;

	for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;

	if(!pre) fa[now] = root;

	else {

		int q = ch[pre][x];

		if(len[pre] + 1 == len[q]) fa[now] = q;

		else {

			int nq = ++tot; fa[nq] = fa[q]; len[nq] = len[pre] + 1;

			memcpy(ch[nq], ch[q], sizeof(ch[q]));

			for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nq;

			fa[q] = fa[now] = nq;

		}

	}

	return las;

}

void Build() {

	for(int i = 1; i <= tot; i++) par[fa[i]].push_back(i);

}

int rt[SS], ls[SS], rs[SS], sum[SS], cnt;

void update(int k) {

	sum[k] = sum[ls[k]] + sum[rs[k]];

}

void Modify(int &k, int l, int r, int p, int v) {

	if(!k) k = ++cnt;

	if(l == r) {sum[k]++; return ;}

	int mid = l + r >> 1;

	if(p <= mid) Modify(ls[k], l, mid, p, v);

	else Modify(rs[k], mid + 1, r, p, v);

	update(k);

}

int Merge(int x, int y) {

	if(!x || !y) return x ^ y;

	int nw = ++cnt;

	if(!ls[x] && !rs[x]) {sum[nw] = sum[x] + sum[y]; return nw;}

	ls[nw] = Merge(ls[x], ls[y]);

	rs[nw] = Merge(rs[x], rs[y]);

	update(nw);

	return nw;

}

int Get(int k, int l, int r) {

	if(!k) return N;

	if(l == r) return l;

	int mid = l + r >> 1;

	if(sum[ls[k]]) return Get(ls[k], l, mid);

	else return Get(rs[k], mid + 1, r);

}

int Query(int k, int l, int r, int ql, int qr) {

	if(!k || (l > r) || (ql > qr)) return 0;

	if(ql <= l && r <= qr)

		return sum[k];

	int mid = l + r >> 1;

	if(ql > mid) return Query(rs[k], mid + 1, r, ql, qr);

	else if(qr <= mid) return Query(ls[k], l, mid, ql, qr);

	else return Query(ls[k], l, mid, ql, qr) + Query(rs[k], mid + 1, r, ql, qr);

}

void dfs(int x) {

	for(auto &to : par[x]) {

		dfs(to);

		rt[x] = Merge(rt[x], rt[to]);

	}

}

void solve() {

	int n = strlen(T + 1), now = root, flag = 0, Lim = 0;

	for(int i = 1; i <= n; i++) {

		int nxt = T[i] - '0';

		if(!ch[now][nxt]) {flag = 1; break;}

		now = ch[now][nxt];

		if(i == n)

			Lim = Get(rt[now], 1, N) - n;//µÚÒ»´Î³öÏÖµÄλÖÃ

	}

	int ans = 0;

	if(flag) ans = N;

	else ans = Lim + n;

	now = root;

	for(int i = 1; i <= n; i++) {

		int nxt = T[i] - '0';

		if(!ch[now][nxt]) break;

		now = ch[now][nxt];

		if(flag) ans += Query(rt[now], 1, N, 1, N);

		else ans += Query(rt[now], 1, N, 1, Lim + i - 1);

	}

	cout << ans << '\n';

}

int main() {

	//freopen("1.in", "r", stdin); freopen("b.out", "w", stdout);

	cin >> N;

	scanf("%s", S + 1);

	for(int i = 1; i <= N; i++)

		Modify(rt[insert(S[i] - '0')], 1, N, i, 1);

	Build();

	dfs(root);

	cin >> M;

	for(int i = 1; i <= M; i++) {

		scanf("%s", T + 1);

		solve();

	}

	return 0;

}

/*

7

1090901

4

0901

87650

109

090

*/

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