SP8093 JZPGYZ

题意

题目链接

Sol

广义后缀自动机板子题。。和BZOJ串那个题很像

首先建出询问串的SAM，然后统计一下每个节点被多少个串包含

最后直接拿询问串上去跑就行了

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
int N, Q;
string s[MAXN], t[MAXN];
int fa[MAXN], len[MAXN], ch[MAXN][26], tim[MAXN], val[MAXN], root = 1, las = 1, tot = 1;
void 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])); 
			fa[q] = fa[now] = nq;
			for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nq;
		}
	}
}
int main() {
	cin >> N >> Q;
	for(int i = 1; i <= N; i++) {
		cin >> s[i]; las = 1;
		for(int j = 0; j < s[i].length(); j++) insert(s[i][j] - 'a');
	}
	for(int i = 1; i <= N; i++) {
		string ns = s[i]; int now = root;
		for(int j = 0; j < ns.length(); j++) {
			int x = ns[j] - 'a';
			now = ch[now][x];
			for(int p = now; p && tim[p] != i; p = fa[p]) 
				tim[p] = i, val[p]++;
		}
	}
	for(int i = 1; i <= Q; i++) {
		string ns; cin >> ns;
		int now = root, flag = 0;
		for(int j = 0; j < ns.length(); j++) {
			int x = ns[j] - 'a';
			if(!ch[now][x]) {flag = 1; break;}
			now = ch[now][x];
		}
		printf("%d
", flag == 1 ? 0 : val[now]);
	}
	return 0;
}