题目
Sol
就是广义(sam)
然后记录下每个状态属于哪些串，开(set)维护
(parent)树上启发式合并一下就好了
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
const int maxn(2e5 + 5);
int trans[26][maxn], fa[maxn], len[maxn], size[maxn], last, tot = 1;
int l[maxn], r[maxn], n, q, t[maxn], id[maxn];
set <int> :: iterator it;
set <int> ed[maxn];
char s[maxn], qs[maxn << 1];
IL void Extend(RG int c){
RG int p = last, np = ++tot;
len[last = np] = len[p] + 1;
while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
if(!p) fa[np] = 1;
else{
RG int q = trans[c][p];
if(len[q] == len[p] + 1) fa[np] = q;
else{
RG int nq = ++tot;
fa[nq] = fa[q], len[nq] = len[p] + 1;
for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
fa[q] = fa[np] = nq;
while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
}
}
}
int main(){
n = Input(), q = Input();
for(RG int i = 1; i <= n; ++i){
l[i] = r[i - 1];
scanf(" %s", s + l[i]);
r[i] = l[i] + strlen(s + l[i]), last = 1;
for(RG int j = l[i]; j < r[i]; ++j) Extend(s[j] - 'a');
}
for(RG int i = 1; i <= tot; ++i) ++t[len[i]];
for(RG int i = 1; i <= tot; ++i) t[i] += t[i - 1];
for(RG int i = 1; i <= tot; ++i) id[t[len[i]]--] = i;
for(RG int i = 1; i <= n; ++i)
for(RG int j = l[i], nw = 1; j < r[i]; ++j){
nw = trans[s[j] - 'a'][nw];
ed[nw].insert(i);
}
for(RG int i = tot; i; --i){
RG int p = id[i];
size[p] = ed[p].size();
if(size[p] > ed[fa[p]].size()) swap(ed[p], ed[fa[p]]);
for(it = ed[p].begin(); it != ed[p].end(); ++it)
ed[fa[p]].insert(*it);
}
for(RG int i = 1; i <= q; ++i){
scanf(" %s", qs);
RG int nw = 1, l = strlen(qs);
for(RG int j = 0; j < l; ++j) nw = trans[qs[j] - 'a'][nw];
printf("%d
", size[nw]);
}
return 0;
}