CodeforcesRound#129(Div.1)E.LittleElephantandStrings

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
#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
#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
#define pil pair
#define pli pair
#define pii pair
#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;}
templateinline T const& MAX(T const &a,T const &b){return a>b?a:b;}
templateinline T const& MIN(T const &a,T const &b){return ainline 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];
vectorv[N];
struct SAM{
int last,cnt;
int ch[N<<1][26],fa[N<<1],l[N<<1];
int sz[N<<1],id[N<<1];
setst[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 }
void dfs(int u)
{
for(int x:son[u])
{
dfs(x);
if(st[id[u]].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 {
now=ch[now][v[i][j]-'a'];
while(now!=1&&sz[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 sam.build(i);
}
sam.cal();
return 0;
}
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