题意:给你一个字符串,找第k大的子字符串.(考虑相同的字符串)
题解:建sam,先预处理出每个节点的出现次数,然后处理出每个节点下面的出现次数,然后在dfs时判断一下往哪边走即可,注意一下num会爆int
//#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<bits/stdc++.h>
#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<int>
#define mod 998244353
#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<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#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;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline 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-7;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=100000+10,inf=0x3f3f3f3f;
struct SAM{
int last,cnt;
int ch[N<<1][26],fa[N<<1],l[N<<1];
ll sz[N<<1],num[N<<1];
int a[N<<1],c[N<<1],vis[N<<1];
void ins(int c){
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;
}
}
sz[np]=1;
}
void topo()
{
for(int i=1;i<=cnt;i++)c[l[i]]++;
for(int i=1;i<=cnt;i++)c[i]+=c[i-1];
for(int i=1;i<=cnt;i++)a[c[l[i]]--]=i;
}
void build(char *s){
int len=strlen(s);
last=cnt=1;
for(int i=0;i<len;i++)ins(s[i]-'a');
topo();
for(int i=cnt;i;i--)sz[fa[a[i]]]+=sz[a[i]];
sz[1]=0;
dfs(1);
// for(int i=1;i<=cnt;i++)
// {
// printf("%d %d ++",i,num[i]);
// for(int j=0;j<26;j++)if(ch[i][j])printf("%d ",ch[i][j]);
// puts("");
// }
}
void dfs(int u)
{
vis[u]=1;
num[u]=sz[u];
for(int i=0;i<26;i++)if(ch[u][i])
{
if(!vis[ch[u][i]])dfs(ch[u][i]);
num[u]+=num[ch[u][i]];
}
}
void cal(int u,int k)
{
if(k<=0)return ;
for(int i=0;i<26;i++)if(ch[u][i])
{
if(num[ch[u][i]]>=k)
{
printf("%c",i+'a');
// printf("%d %d %d %c %d
",u,ch[u][i],num[ch[u][i]],i+'a',k);
cal(ch[u][i],k-sz[ch[u][i]]);
return ;
}
else k-=num[ch[u][i]];
}
}
}sam;
char s[N];
int main()
{
int k;
scanf("%s%d",s,&k);
sam.build(s);
if(sam.num[1]<k)puts("No such line.");
else sam.cal(1,k);
return 0;
}
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