BZOJ3413: 匹配(后缀自动机 线段树合并)

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
*/