cf204E&bzoj3277&bzoj3473

题解:把所有串连起来做一次sa 对于每个位置的长度做一次二分，对于每个位置找到最长的合适长度，用可持久化结构在维护，时间复杂度nlogn^2；

#include <bits/stdc++.h>
#define ll long long
const int MAXN=2e5+10;
using namespace std;
typedef struct node{
    int l,r,sum;
}node;
node d[MAXN*25];
int cnt;
char str[MAXN];char str1[MAXN];
int sa[MAXN],txt[MAXN],rank1[MAXN],rank2[MAXN],td[MAXN],t1[MAXN],t2[MAXN];
int n,k;
bool cmp(int f[],int w,int m,int k){return f[w]==f[m]&&f[w+k]==f[m+k];}
void Sa(){
    int len=strlen(str);int m=128;
    int *td=t1;int *rank1=t2;
    for(int i=0;i<m;i++)txt[i]=0;
    for(int i=0;i<len;i++)rank1[i]=str[i],txt[str[i]]++;
    for(int i=1;i<m;i++)txt[i]+=txt[i-1];
    for(int i=len-1;i>=0;i--)sa[--txt[str[i]]]=i;
    for(int k=1;k<=len;k*=2){
	int p=0;
	for(int i=len-k;i<len;i++)td[p++]=i;
	for(int i=0;i<len;i++)if(sa[i]>=k)td[p++]=sa[i]-k;
	for(int i=0;i<m;i++)txt[i]=0;
	for(int i=0;i<len;i++)txt[rank1[i]]++;
	for(int i=1;i<m;i++)txt[i]+=txt[i-1];
	for(int i=len-1;i>=0;i--)sa[--txt[rank1[td[i]]]]=td[i];
	swap(rank1,td);rank1[sa[0]]=0;p=1;
	for(int i=1;i<len;i++)rank1[sa[i]]=cmp(td,sa[i],sa[i-1],k)?p-1:p++;
	if(p==len)return ;
	m=p;
    }
}
int h[MAXN],H[MAXN];
void hh(){
    int len=strlen(str);
    for(int i=0;i<len;i++)rank2[sa[i]]=i;
    memset(H,0,sizeof(H));
    for(int i=0;i<len;i++){
	if(rank2[i]==0)continue;
	int t=sa[rank2[i]-1];int w=i;int k;
	if(i==0||H[i-1]<=1)k=0;
	else k=H[i-1]-1,t+=k,w+=k;
	while(t<len&&w<len){
	    if(str[t]==str[w])k++;
	    else break;
	    t++;w++;
	}
	H[i]=k;h[rank2[i]]=k;
    }
}
int dp[MAXN][21],mu[21],ma[MAXN];
void st(int len){
    mu[0]=1;
    for(int i=1;i<21;i++)mu[i]=mu[i-1]<<1;
    ma[0]=-1;
    for(int i=1;i<len;i++){
	if((i&(i-1))==0)ma[i]=ma[i-1]+1;
	else ma[i]=ma[i-1];
    }
   // for(int i=1;i<len;i++)
    for(int i=1;i<len;i++)dp[i][0]=h[i];
    for(int j=1;mu[j]<=len;j++){
	for(int i=1;i+mu[j]<=len;i++){
	    dp[i][j]=min(dp[i][j-1],dp[i+mu[j-1]][j-1]);
	}
    }
}
int rmq(int l,int r){
    int k=ma[r-l+1];
  //  cout<<l<<" "<<r<<" "<<k<<" "<<dp[l][k]<<" "<<dp[r-mu[k]+1][k]<<endl;
    return min(dp[l][k],dp[r-mu[k]+1][k]);
}
void update(int &x,int y,int l,int r,int t,int vul){
    x=++cnt;d[x]=d[y];d[x].sum+=vul;
    if(l==r)return ;
    int mid=(l+r)>>1;
    if(t<=mid)update(d[x].l,d[y].l,l,mid,t,vul);
    else update(d[x].r,d[y].r,mid+1,r,t,vul);
}
int ans;
void querty(int x,int l,int r,int ql,int qr){
    if(ql<=l&&r<=qr){ans+=d[x].sum;return ;}
    int mid=(l+r)>>1;
    if(ql<=mid)querty(d[x].l,l,mid,ql,qr);
    if(qr>mid)querty(d[x].r,mid+1,r,ql,qr);
}
int vis[MAXN],length[MAXN],pos[MAXN],rt[MAXN];
ll sum[MAXN];
bool check(int t,int p,int len){
    int l=1;int r=p;int ans1=p+1;
    while(l<=r){
	int mid=(l+r)>>1;
//	cout<<mid<<" "<<p<<"====="<<rmq(mid,p)<<endl;
	if(rmq(mid,p)>=t)ans1=mid,r=mid-1;
	else l=mid+1;
    }
    //if(ans1!=p+1)ans1--;
    ans1--;
    //if(ans1<0)ans1=0;
    l=p+1;r=len-1;int ans2=p;
    while(l<=r){
	int mid=(l+r)>>1;
//	cout<<p+1<<"===="<<mid<<" "<<rmq(p+1,mid)<<endl;
	if(rmq(p+1,mid)>=t)ans2=mid,l=mid+1;
	else r=mid-1;
    }
  //  cout<<ans1<<"----===="<<ans2<<" "<<t<<endl;
    ans=0;querty(rt[ans2],1,len,ans1,ans2);
  //  cout<<ans<<endl;
    if(ans>=k)return true;
    return false;
}
void solve(){
    int len=strlen(str);
    for(int i=1;i<len;i++){
	if(!vis[sa[i]])continue;
	int t1=pos[sa[i]];int l=0;int r=length[vis[sa[i]]]-t1;
	int ans1=0;
//	cout<<i<<" "<<l<<" "<<r<<" "<<vis[sa[i]]<<endl;
	while(l<=r){
	    int mid=(l+r)>>1;
	    if(check(mid,i,len))ans1=mid,l=mid+1;
	    else r=mid-1;
	}
//	cout<<ans1<<endl;
	sum[vis[sa[i]]]+=1ll*ans1;
    }
}
int pre[MAXN];
int main(){
    scanf("%d%d",&n,&k);
    int len=0;
    for(int i=1;i<=n;i++){
	scanf("%s",str1);int len1=strlen(str1);length[i]=len1;
	for(int j=0;j<len1;j++)vis[len]=i,pos[len]=j,str[len++]=str1[j];
	vis[len]=0,pos[len]=0;str[len++]='$';
    }
    Sa();hh();st(len);
 //   cout<<len<<endl;
 //   cout<<str<<endl;
//    for(int i=1;i<len;i++)cout<<h[i]<<" ";
 //   cout<<endl;
    for(int i=1;i<len;i++){
	if(!vis[sa[i]]){rt[i]=rt[i-1];continue;}
//	cout<<pre[vis[sa[i]]]<<" ";
	update(rt[i],rt[i-1],1,len,i,1);
	if(pre[vis[sa[i]]])update(rt[i],rt[i],1,len,pre[vis[sa[i]]],-1);
	pre[vis[sa[i]]]=i;
    }
  //  cout<<endl;
    solve();
    for(int i=1;i<=n;i++)printf("%lld ",sum[i]);
    puts("");
}

　　　　　　　　　　　　E. Little Elephant and Strings

　　　　　　time limit per test

　　　　　　3 seconds

　　　　　　memory limit per test 256 megabytes

　　　　　　input

　　　　　　standard input

　　　　　　output

　　　　　　standard output

The Little Elephant loves strings very much.

He has an array a from n strings, consisting of lowercase English letters. Let's number the elements of the array from 1 to n, then let's denote the element number i as a_i. For each string a_i (1 ≤ i ≤ n) the Little Elephant wants to find the number of pairs of integers l and r (1 ≤ l ≤ r ≤ |a_i|) such that substring a_i[l... r] is a substring to at least k strings from array a (including the i-th string).

Help the Little Elephant solve this problem.

If you are not familiar with the basic notation in string problems, you can find the corresponding definitions in the notes.

Input

The first line contains two space-separated integers — n and k (1 ≤ n, k ≤ 10⁵). Next n lines contain array a. The i-th line contains a non-empty string a_i, consisting of lowercase English letter. The total length of all strings a_i does not exceed 10⁵.

Output

On a single line print n space-separated integers — the i-th number is the answer for string a_i.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

Copy

3 1
abc
a
ab

Output

Copy

6 1 3 

Input

Copy

7 4
rubik
furik
abab
baba
aaabbbababa
abababababa
zero

Output

Copy

1 0 9 9 21 30 0 

Note

Let's assume that you are given string a = a₁a₂... a_|a|, then let's denote the string's length as |a| and the string's i-th character as a_i.

A substring a[l... r] (1 ≤ l ≤ r ≤ |a|) of string a is string a_la_l + 1... a_r.

String a is a substring of string b, if there exists such pair of integers l and r (1 ≤ l ≤ r ≤ |b|), that b[l... r] = a.