# 洛谷P2852 [USACO06DEC]牛奶模式Milk Patterns

2018-02-04 18:33:19来源:cnblogs.com作者:自为风月马前卒人点击

## 题目描述

Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation, he discovered that although he can't predict the quality of milk from one day to the next, there are some regular patterns in the daily milk quality.

To perform a rigorous study, he has invented a complex classification scheme by which each milk sample is recorded as an integer between 0 and 1,000,000 inclusive, and has recorded data from a single cow over N (1 ≤ N ≤ 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 ≤ K ≤ N) times. This may include overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example.

Help Farmer John by finding the longest repeating subsequence in the sequence of samples. It is guaranteed that at least one subsequence is repeated at least K times.

## 输入输出格式

Line 1: Two space-separated integers: N and K

Lines 2..N+1: N integers, one per line, the quality of the milk on day i appears on the ith line.

Line 1: One integer, the length of the longest pattern which occurs at least K times

## 输入输出样例

`8 212323231`

`4`

`#include<cstdio>#include<cstring>#include<queue>using namespace std;const int MAXN=200001,INF=2*1e9+10;inline char nc(){    static char buf[MAXN],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++;}inline int read(){    char c=nc();int x=0,f=1;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();}    return x*f;}int N,M,K;int a[MAXN],sa[MAXN],tax[MAXN],tp[MAXN],rak[MAXN],Height[MAXN];inline void Qsort(){    for(int i=1;i<=M;i++) tax[i]=0;    for(int i=1;i<=N;i++) tax[rak[i]]++;    for(int i=1;i<=M;i++) tax[i]+=tax[i-1];    for(int i=N;i>=1;i--) sa[ tax[rak[tp[i]]]-- ] = tp[i];}int check(int val){    int cur=0;    for(int i=1;i<=N;i++)    {        if(Height[i]<val) cur=0;        else cur++;        if(cur>=K-1) return 1;    }    return 0;}inline void Ssort(){    M=10001;    for(int i=1;i<=N;i++) rak[i]=a[i],tp[i]=i;Qsort();    for(int w=1,p=1; p<N ; M=p,w<<=1)    {        p=0;        for(int i=N-w+1;i<=N;i++) tp[++p]=i;        for(int i=1;i<=N;i++) if(sa[i]>w) tp[++p]=sa[i]-w;        Qsort();swap(rak,tp);rak[sa[1]]=p=1;        for(int i=2;i<=N;i++) rak[sa[i]]=(tp[sa[i]]==tp[sa[i-1]]&&tp[sa[i]+w]==tp[sa[i-1]+w])?p:++p;    }    int j,k=0;//    for(int i=1;i<=N;i++) printf("%d ",sa[i]);printf("/n");    for(int i=1;i<=N;Height[rak[i++]]=k)        for( k=(k?k-1:k),j=sa[rak[i]-1];a[i+k]==a[j+k];++k);//    for(int i=1;i<=N;i++) printf("%d ",Height[i]);    int l=1,r=N,ans=0;    while(l<=r)    {        int mid=(l+r)>>1;        if(check(mid)) ans=mid,l=mid+1;        else r=mid-1;    }    printf("%d",ans);}int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else    #endif    N=read();K=read();    for(int i=1;i<=N;i++) a[i]=read();    Ssort();    return  0;}`