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.
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.
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.
3 1
abc
a
ab
6 1 3
7 4
rubik
furik
abab
baba
aaabbbababa
abababababa
zero
1 0 9 9 21 30 0
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.
这道题很好啊……
本来要用后缀数组，现在我用的是后缀自动机。用set维护经过的字串有哪些，建成这棵trie的SAM后，连边fa[i]->i，然后DFS，把set启发式合并一下，要的信息就是size，处理答案时先跳（必定可行），然后size<k就不停跳fa，直到size>=k为止，这时匹配的答案加入计数。
1 #include <iostream> 2 #include <cstring> 3 #include <cstring> 4 #include <cstdio> 5 #include <set> 6 using namespace std; 7 const int N=200010; 8 int fa[N],ch[N][27],len[N]; 9 string s[N];set<int>t[N]; 10 int tr[N][26],last[N]; 11 int lst,cnt,n,k,tot; 12 void Insert(int c){ 13 int p=lst,np=lst=++cnt;len[np]=len[p]+1; 14 while(p&&ch[p][c]==0)ch[p][c]=np,p=fa[p]; 15 if(!p)fa[np]=1; 16 else{ 17 int q=ch[p][c],nq; 18 if(len[q]==len[p]+1) 19 fa[np]=q; 20 else{ 21 len[nq=++cnt]=len[p]+1; 22 fa[nq]=fa[q];fa[q]=fa[np]=nq; 23 memcpy(ch[nq],ch[q],sizeof(ch[q])); 24 while(ch[p][c]==q)ch[p][c]=nq,p=fa[p]; 25 } 26 } 27 } 28 int cntE,fir[N],to[N],nxt[N]; 29 void addedge(int a,int b){ 30 nxt[++cntE]=fir[a]; 31 to[fir[a]=cntE]=b; 32 } 33 void Initialization(){ 34 memset(fir,0,sizeof(fir)); 35 lst=cnt=1;last[cntE=tot=0]=1; 36 } 37 set<int>Merge(set<int>a,set<int>b){ 38 if(a.size()<b.size())swap(a,b); 39 a.insert(b.begin(),b.end()); 40 delete &b;return a; 41 } 42 void DFS(int x){ 43 for(int i=fir[x],g;i;i=nxt[i]) 44 DFS(g=to[i]),t[x]=Merge(t[x],t[g]); 45 } 46 47 int main(){ 48 Initialization(); 49 scanf("%d%d",&n,&k); 50 for(int i=1;i<=n;i++){ 51 cin>>s[i]; 52 for(int j=0,p=0,c;j<s[i].size();j++){ 53 c=s[i][j]-'a';lst=last[p]; 54 if(tr[p][c])p=tr[p][c],t[last[p]].insert(i); 55 else Insert(c),t[last[p=tr[p][c]=++tot]=lst].insert(i); 56 } 57 } 58 for(int i=2;i<=cnt;i++) 59 addedge(fa[i],i);DFS(1); 60 for(int i=1;i<=n;i++){ 61 long long ans=0; 62 for(int j=0,p=1,c,l=0;j<s[i].size();j++){ 63 c=s[i][j]-'a'; 64 p=ch[p][c];l+=1; 65 while(p!=1&&t[p].size()<k) 66 p=fa[p],l=len[p]; 67 ans+=l; 68 } 69 printf("%I64d ",ans); 70 } 71 printf(" "); 72 return 0; 73 }