传送门

题意：

思路：

这个只能搞$fail$树了，让后跑一遍即可。

重复的字符串记一个，让后其他的都跟他一个答案即可。

// Problem: P3796 【模板】AC自动机（加强版）
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3796
// Memory Limit: 128 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
int tr[N][26],idx,ed[N],fail[N];
int cnt[N],pos[N],ans[N];
string s[N],ss;
vector<int>v[N];int newnode() {int u=++idx;for(int i=0;i<26;i++) tr[u][i]=0;return u; 
}void insert(int id,string s) {int n=s.length(),p=0;for(int i=0;i<n;i++) {int u=s[i]-'a';if(!tr[p][u]) tr[p][u]=newnode();p=tr[p][u];}ed[p]=id;
}void get_fail() {queue<int>q;for(int i=0;i<26;i++) if(tr[0][i]) q.push(tr[0][i]);while(q.size()) {int u=q.front(); q.pop();for(int i=0;i<26;i++) {int p=tr[u][i];if(p) fail[p]=tr[fail[u]][i],q.push(p);else tr[u][i]=tr[fail[u]][i];}}
}void dfs(int u) {for(auto x:v[u]) dfs(x),cnt[u]+=cnt[x];
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);scanf("%d",&n);idx=0; for(int i=0;i<26;i++) tr[0][i]=0;map<string,int>mp;for(int i=1;i<=n;i++) {cin>>s[i]; if(!mp.count(s[i])) mp[s[i]]=i;insert(mp[s[i]],s[i]);}get_fail();cin>>ss;for(int i=1;i<=idx;i++) v[fail[i]].pb(i);int len=ss.length();int p=0;for(int i=0;i<len;i++) {int u=ss[i]-'a';p=tr[p][u];cnt[p]++;}dfs(0);for(int i=1;i<=idx;i++) if(ed[i]) ans[ed[i]]+=cnt[i];for(int i=1;i<=n;i++) printf("%d\n",ans[mp[s[i]]]); return 0;
}
/**/