http://www.elijahqi.win/archives/1223
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 ai. For each string ai (1 ≤ i ≤ n) the Little Elephant wants to find the number of pairs of integers l and r (1 ≤ l ≤ r ≤ |ai|) such that substring ai[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 ai, consisting of lowercase English letter. The total length of all strings ai does not exceed 105.
Output
On a single line print n space-separated integers — the i-th number is the answer for string ai.
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
3 1
abc
a
ab
Output
6 1 3
Input
7 4
rubik
furik
abab
baba
aaabbbababa
abababababa
zero
Output
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 ai.
A substring a[l… r] (1 ≤ l ≤ r ≤ |a|) of string a is string alal + 1… ar.
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.
大概是原题bzoj3473http://www.elijahqi.win/2017/08/26/bzoj3473/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 220000
int st[N][20],height[N],rank[N<<1],rank1[N],s[N],L[N],Log[N],n,kk,cnt[N],bl[N>>1],sa[N],count[N],tmp[N];
char a[N];
inline int lcp(int x,int y){
x++;int t=Log[y-x+1];
return std::min(st[x][t],st[y-(1<<t)+1][t]);
}
struct node{int st,ed;}data[N>>1];
inline bool check(int st,int len){
int l1,r1;
if (height[st]<len) l1=st;else{
int l=1,r=st-1;
while (l<=r){
int mid=l+r>>1;
if (lcp(mid,st)>=len) r=mid-1;else l=mid+1;
}l1=l;
}
if (height[st+1]<len) r1=st;else{
int l=st+1,r=n;
while (l<=r){
int mid=l+r>>1;
if(lcp(st,mid)>=len) l=mid+1;else r=mid-1;
}r1=r;
}
return L[r1]>=l1;
}
int main(){
freopen("cf.in","r",stdin);
scanf("%d%d",&n,&kk);int len=0,m=30;
for (int i=1;i<=n;++i){
scanf("%s",a+1);int len1=strlen(a+1);data[i].st=len+1;
for (int j=len+1;j<=len+len1;++j) s[j]=a[j-len]-'a'+1,bl[j]=i;data[i].ed=len+len1;len+=len1;s[++len]=++m;
}int nn=n;n=len;
for (int i=1;i<=n;++i) count[s[i]]=1;
for (int i=1;i<=m;++i) count[i]+=count[i-1];
for (int i=1;i<=n;++i) rank[i]=count[s[i]];int k=0;
for (int p=1;k!=n;p<<=1,m=k){
for (int i=1;i<=m;++i) count[i]=0;
for (int i=1;i<=n;++i) count[rank[i+p]]++;
for (int i=1;i<=m;++i) count[i]+=count[i-1];
for (int i=n;i>=1;--i) tmp[count[rank[i+p]]--]=i;
for (int i=1;i<=m;++i) count[i]=0;
for (int i=1;i<=n;++i) count[rank[i]]++;
for (int i=1;i<=m;++i) count[i]+=count[i-1];
for (int i=n;i>=1;--i) sa[count[rank[tmp[i]]]--]=tmp[i];
memcpy(rank1,rank,sizeof(rank)>>1);
rank[sa[1]]=k=1;
for (int i=2;i<=n;++i){
if (rank1[sa[i]]!=rank1[sa[i-1]]||rank1[sa[i]+p]!=rank1[sa[i-1]+p]) ++k;
rank[sa[i]]=k;
}
}k=0;
for (int i=1;i<=n;++i){
if (rank[i]==1) continue;
k=k==0?0:k-1;
while (s[i+k]==s[sa[rank[i]-1]+k]) ++k;
height[rank[i]]=k;
}Log[0]=-1;for (int i=1;i<=n;++i) Log[i]=Log[i>>1]+1;
for (int i=1;i<=n;++i) st[i][0]=height[i];
for (int j=1;j<=Log[n];++j)
for (int i=1;i+(1<<j)<=n+1;++i) st[i][j]=std::min(st[i][j-1],st[i+(1<<(j-1))][j-1]);
int num=0;
// for (int i=1;i<=n;++i) printf("%d ",height[i]);printf("\n");
for (int i=1,p=1;i<=n;++i){
if (!bl[sa[i]]) break;
if (!cnt[bl[sa[i]]]) num++;cnt[bl[sa[i]]]++;
while (num>kk||num==kk&&cnt[bl[sa[p]]]>1){
--cnt[bl[sa[p]]];if (!cnt[bl[sa[p++]]]) num--;
}
if (num>=kk) L[i]=p;else L[i]=-1;
}
for (int i=1;i<=nn;++i){
long long ans=0;k=0;
for (int j=data[i].st;j<=data[i].ed;++j){
k=k==0?0:k-1;
while (j+k<=data[i].ed&&check(rank[j],k+1))++k;ans+=k;
}
printf("%lld ",ans);
}
return 0;
}