如果颜色数 (geq) k 就可以 (cnt_u += len_u - len_{fa_u}) 也就是这个点包含的字符串数量。
然后 (cnt_u) 表示 u 节点对应的字符串所包含的子串的贡献,这样就可以算完了。
就很简单了。
// clang-format off
// powered by c++11
// by Isaunoya
#include<bits/stdc++.h>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define Rep(i,x,y) for(register int i=(x);i>=(y);--i)
using namespace std;using db=double;using ll=long long;
using uint=unsigned int;using ull=unsigned long long;
using pii=pair<int,int>;
#define Tp template
#define fir first
#define sec second
Tp<class T>void cmax(T&x,const T&y){if(x<y)x=y;}Tp<class T>void cmin(T&x,const T&y){if(x>y)x=y;}
#define all(v) v.begin(),v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T>void sort(vector<T>&v){sort(all(v));}Tp<class T>void reverse(vector<T>&v){reverse(all(v));}
Tp<class T>void unique(vector<T>&v){sort(all(v)),v.erase(unique(all(v)),v.end());}inline void reverse(string&s){reverse(s.begin(),s.end());}
const int SZ=1<<23|233;
struct FILEIN{char qwq[SZ],*S=qwq,*T=qwq,ch;
#ifdef __WIN64
#define GETC getchar
#else
inline char GETC(){return(S==T)&&(T=(S=qwq)+fread(qwq,1,SZ,stdin),S==T)?EOF:*S++;}
#endif
inline FILEIN&operator>>(char&c){while(isspace(c=GETC()));return*this;}inline FILEIN&operator>>(string&s){s.clear();while(isspace(ch=GETC()));if(!~ch)return*this;s=ch;while(!isspace(ch=GETC())&&~ch)s+=ch;return*this;}
inline FILEIN&operator>>(char*str){char*cur=str;while(*cur)*cur++=0;cur=str;while(isspace(ch=GETC()));if(!~ch)return*this;*cur=ch;while(!isspace(ch=GETC())&&~ch)*++cur=ch;*++cur=0;return*this;}
Tp<class T>inline void read(T&x){bool f=0;while((ch=GETC())<48&&~ch)f^=(ch==45);x=~ch?(ch^48):0;while((ch=GETC())>47)x=x*10+(ch^48);x=f?-x:x;}
inline FILEIN&operator>>(int&x){return read(x),*this;}inline FILEIN&operator>>(ll&x){return read(x),*this;}inline FILEIN&operator>>(uint&x){return read(x),*this;}inline FILEIN&operator>>(ull&x){return read(x),*this;}
inline FILEIN&operator>>(double&x){read(x);bool f=x<0;x=f?-x:x;if(ch^'.')return*this;double d=0.1;while((ch=GETC())>47)x+=d*(ch^48),d*=.1;return x=f?-x:x,*this;}
}in;
struct FILEOUT{const static int LIMIT=1<<22;char quq[SZ],ST[233];int sz,O,pw[233];
FILEOUT(){set(7);rep(i,pw[0]=1,9)pw[i]=pw[i-1]*10;}~FILEOUT(){flush();}
inline void flush(){fwrite(quq,1,O,stdout),fflush(stdout),O=0;}
inline FILEOUT&operator<<(char c){return quq[O++]=c,*this;}inline FILEOUT&operator<<(string str){if(O>LIMIT)flush();for(char c:str)quq[O++]=c;return*this;}
inline FILEOUT&operator<<(char*str){if(O>LIMIT)flush();char*cur=str;while(*cur)quq[O++]=(*cur++);return*this;}
Tp<class T>void write(T x){if(O>LIMIT)flush();if(x<0){quq[O++]=45;x=-x;}do{ST[++sz]=x%10^48;x/=10;}while(x);while(sz)quq[O++]=ST[sz--];}
inline FILEOUT&operator<<(int x){return write(x),*this;}inline FILEOUT&operator<<(ll x){return write(x),*this;}inline FILEOUT&operator<<(uint x){return write(x),*this;}inline FILEOUT&operator<<(ull x){return write(x),*this;}
int len,lft,rig;void set(int l){len=l;}inline FILEOUT&operator<<(double x){bool f=x<0;x=f?-x:x,lft=x,rig=1.*(x-lft)*pw[len];return write(f?-lft:lft),quq[O++]='.',write(rig),*this;}
}out;
struct Math{
vector<int>fac,inv;int mod;
void set(int n,int Mod){fac.resize(n+1),inv.resize(n+1),mod=Mod;rep(i,fac[0]=1,n)fac[i]=fac[i-1]*i%mod;inv[n]=qpow(fac[n],mod-2);Rep(i,n-1,0)inv[i]=inv[i+1]*(i+1)%mod;}
int qpow(int x,int y){int ans=1;for(;y;y>>=1,x=x*x%mod)if(y&1)ans=ans*x%mod;return ans;}int C(int n,int m){if(n<0||m<0||n<m)return 0;return fac[n]*inv[m]%mod*inv[n-m]%mod;}
int gcd(int x,int y){return!y?x:gcd(y,x%y);}int lcm(int x,int y){return x*y/gcd(x,y);}
}math;
// clang-format on
int n, k;
const int maxn = 2e5 + 52;
set<int> s[maxn];
struct suffix_auto_maton {
int las, cnt;
suffix_auto_maton() { las = cnt = 1; }
int ch[maxn][26], fa[maxn], len[maxn];
void ins(int c, int id) {
int p = las, np = las = ++cnt;
len[np] = len[p] + 1, s[np].insert(id);
for (; p && !ch[p][c]; p = fa[p]) ch[p][c] = np;
if (!p) {
fa[np] = 1;
} else {
int q = ch[p][c];
if (len[q] == len[p] + 1) {
fa[np] = q;
} else {
int nq = ++cnt;
memcpy(ch[nq], ch[q], sizeof(ch[q]));
len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
for (; p && ch[p][c] == q; p = fa[p]) ch[p][c] = nq;
}
}
}
void ins(string s, int id) {
las = 1;
for (char c : s) {
ins(c - 'a', id);
}
}
} sam;
int col[maxn];
void merge(int u, int v) {
if (s[u].size() < s[v].size()) swap(s[u], s[v]);
for (int x : s[v]) s[u].insert(x);
s[v].clear();
}
vector<int> g[maxn];
void dfs(int u) {
for (int v : g[u]) {
dfs(v);
merge(u, v);
}
col[u] = (int)s[u].size();
}
ll cnt[maxn];
void dfs2(int u) {
if (col[u] >= k) cnt[u] += sam.len[u] - sam.len[sam.fa[u]];
for (int v : g[u]) {
cnt[v] = cnt[u];
dfs2(v);
}
}
string str[maxn];
signed main() {
// code begin.
in >> n >> k;
rep(i, 1, n) {
in >> str[i];
sam.ins(str[i], i);
}
rep(i, 2, sam.cnt) g[sam.fa[i]].pb(i);
dfs(1);
dfs2(1);
rep(i, 1, n) {
int p = 1;
ll ans = 0;
for (char x : str[i]) {
int c = x - 'a';
p = sam.ch[p][c];
ans += cnt[p];
}
out << ans << ' ';
}
return 0;
// code end.
}