題目：http://poj.org/problem?id=3261
題意：給你一個序列，求序列裏重複出現至少K次的最長子串
分析：這題如果學過後綴數組的話，那就是模版題了，直接構造一個後綴數組，然後枚舉i，詢問[ i, i+k-1 ] 的最長公共前綴就行，i表示排在第i位的後綴
代碼：
/** head files*/
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;
/** some operate*/
#define PB push_back
#define MP make_pair
#define REP(i,n) for(i=0;i<(n);++i)
#define UPTO(i,l,h) for(i=(l);i<=(h);++i)
#define DOWN(i,h,l) for(i=(h);i>=(l);--i)
#define MSET(arr,val) memset(arr,val,sizeof(arr))
#define MAX3(a,b,c) max(a,max(b,c))
#define MAX4(a,b,c,d) max(max(a,b),max(c,d))
#define MIN3(a,b,c) min(a,min(b,c))
#define MIN4(a,b,c,d) min(min(a,b),min(c,d))
/** some const*/
#define N 222222
#define M 222222
#define PI acos(-1.0)
#define oo 1111111111
/** some alias*/
typedef long long ll;
/** Global variables*/
/** some template names, just push ctrl+j to get it in*/
//manacher 求最長迴文子串
//pqueue 優先隊列
//combk n元素序列的第m小的組合和
//pmatrix n個點的最大子矩陣
//suffixarray 後綴數組
template <typename T, int LEN>
struct suffixarray
{
int str[LEN*3],sa[LEN*3];
int rank[LEN],height[LEN];
int id[LEN];
int best[LEN][20];
int len;
bool equal(int *str, int a, int b)
{
return str[a]==str[b]&&str[a+1]==str[b+1]&&str[a+2]==str[b+2];
}
bool cmp3(int *str, int *nstr, int a, int b)
{
if(str[a]!=str[b])return str[a]<str[b];
if(str[a+1]!=str[b+1])return str[a+1]<str[b+1];
return nstr[a+b%3]<nstr[b+b%3];
}
void radixsort(int *str, int *sa, int *res, int n, int m)
{
int i;
REP(i,m)id[i]=0;
REP(i,n)++id[str[sa[i]]];
REP(i,m)id[i+1]+=id[i];
DOWN(i,n-1,0)res[--id[str[sa[i]]]]=sa[i];
}
void dc3(int *str, int *sa, int n, int m)
{
#define F(x) ((x)/3+((x)%3==1?0:one))
#define G(x) ((x)<one?(x)*3+1:((x)-one)*3+2)
int *nstr=str+n, *nsa=sa+n, *tmpa=rank, *tmpb=height;
int i,j,k,len=0,num=0,zero=0,one=(n+1)/3;
REP(i,n)if(i%3)tmpa[len++]=i;
str[n]=str[n+1]=0;
radixsort(str+2, tmpa, tmpb, len, m);
radixsort(str+1, tmpb, tmpa, len, m);
radixsort(str+0, tmpa, tmpb, len, m);
nstr[F(tmpb[0])]=num++;
UPTO(i,1,len-1)
nstr[F(tmpb[i])]=equal(str,tmpb[i-1],tmpb[i])?num-1:num++;
if(num<len)dc3(nstr,nsa,len,num);
else REP(i,len)nsa[nstr[i]]=i;
if(n%3==1)tmpa[zero++]=n-1;
REP(i,len)if(nsa[i]<one)tmpa[zero++]=nsa[i]*3;
radixsort(str, tmpa, tmpb, zero, m);
REP(i,len)tmpa[nsa[i]=G(nsa[i])]=i;
i=j=0;
REP(k,n)
if(j>=len||(i<zero&&cmp3(str,tmpa,tmpb[i],nsa[j])))sa[k]=tmpb[i++];
else sa[k]=nsa[j++];
}
void initSA(T *s, int n,int m)
{
int i,j,k=0;
str[len=n]=0;
REP(i,n)str[i]=s[i];
dc3(str,sa,n+1,m);
REP(i,n)sa[i]=sa[i+1];
REP(i,n)rank[sa[i]]=i;
REP(i,n)
{
if(k)--k;
if(rank[i])for(j=sa[rank[i]-1];str[i+k]==str[j+k];++k);
else k=0;
height[rank[i]]=k;
}
}
void initRMQ()
{
int i,j;
REP(i,len)best[i][0]=height[i];
for(j=1;(1<<j)-1<len;++j)
for(i=0;i+(1<<j)-1<len;++i)
best[i][j]=min(best[i][j-1],best[i+(1<<(j-1))][j-1]);
}
int RMQ(int l, int r)
{
int k=0;
while(l+(1<<k)-1+1<r-(1<<k)+1)++k;
return min(best[l][k],best[r-(1<<k)+1][k]);
}
int LCPSA(int a, int b)
{
if(a==b)return len-sa[a];
if(++a>b)swap(a,b);
return RMQ(a,b);
}
};
suffixarray<int,N> msa;
map<int ,int > mymap;
int s[N];
int main()
{
int i,n,m,k,ans;
while(~scanf("%d%d",&n,&k))
{
mymap.clear();
m=1;
REP(i,n)
{
scanf("%d",&s[i]);
if(mymap.find(s[i])==mymap.end())
mymap[s[i]]=m++;
s[i]=mymap[s[i]];
}
msa.initSA(s,n,m);
msa.initRMQ();
ans=0;
REP(i,n)
if(i+k-1<n)
ans=max(ans,msa.LCPSA(i,i+k-1));
printf("%d",ans);
}
return 0;
}