题意:有n个串,询问每个串有多少子串在n个串中出现了至少k次.
题解:sam,每个节点开一个set维护该节点的字符串有哪几个串,启发式合并set,然后在sam上走一遍该串,对于每个可行的串,所有的fail都是可行的直接加上len,不可行就往fail上跳.
被for(int i=0;s[i];i++)给骚死了,一直tle....

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
//#define ls l,m,rt<<1
//#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=1000000+10,inf=0x3f3f3f3f;

int n,k;
char s[N];
vector<char>v[N];
struct SAM{
    int last,cnt;
    int ch[N<<1][26],fa[N<<1],l[N<<1];
    int sz[N<<1],id[N<<1];
    set<int>st[N<<1];
    vi son[N<<1];
    SAM(){cnt=1;}
    void ins(int c){
        if(ch[last][c])
        {
            int p=last,q=ch[last][c];
            if(l[q]==l[p]+1)last=q;
            else
            {
                int nq=++cnt;l[nq]=l[p]+1;
                memcpy(ch[nq],ch[q],sizeof ch[q]);
                fa[nq]=fa[q];fa[q]=last=nq;
                for(;ch[p][c]==q;p=fa[p])ch[p][c]=nq;
            }
            return ;
        }
        int p=last,np=++cnt;last=np;l[np]=l[p]+1;
        for(;p&&!ch[p][c];p=fa[p])ch[p][c]=np;
        if(!p)fa[np]=1;
        else
        {
            int q=ch[p][c];
            if(l[p]+1==l[q])fa[np]=q;
            else
            {
                int nq=++cnt;l[nq]=l[p]+1;
                memcpy(ch[nq],ch[q],sizeof(ch[q]));
                fa[nq]=fa[q];fa[q]=fa[np]=nq;
                for(;ch[p][c]==q;p=fa[p])ch[p][c]=nq;
            }
        }
    }
    void build(int id)
    {
        last=1;int len=strlen(s);
        for(int i=0;i<len;i++)ins(s[i]-'a'),st[last].insert(id);
    }
    void dfs(int u)
    {
        for(int x:son[u])
        {
            dfs(x);
            if(st[id[u]].size()<st[id[x]].size())
            {
                for(int p:st[id[u]])st[id[x]].insert(p);
                st[id[u]].clear();id[u]=id[x];
            }
            else
            {
                for(int p:st[id[x]])st[id[u]].insert(p);
                st[id[x]].clear();
            }
        }
        sz[u]=st[id[u]].size();
    }
    void cal()
    {
        for(int i=2;i<=cnt;i++)son[fa[i]].pb(i),id[i]=i;
        id[1]=1;
        dfs(1);
        for(int i=1;i<=n;i++)
        {
            int now=1;ll ans=0;
            for(int j=0;j<v[i].size();j++)
            {
                now=ch[now][v[i][j]-'a'];
                while(now!=1&&sz[now]<k)now=fa[now];
                ans+=l[now];
            }
            printf("%lld ",ans);
        }
        puts("");
    }
}sam;
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%s",s);
        int n=strlen(s);
        for(int j=0;j<n;j++)v[i].pb(s[j]);
        sam.build(i);
    }
    sam.cal();
    return 0;
}
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