POJ 3261 Milk Patterns （后缀数组或HASH，4级）

G - Milk Patterns

Crawling in process...Crawling failedTime Limit:5000MS    Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

SubmitStatus

Appoint description:System Crawler (2013-08-01)

Description

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 overN (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 leastK times.

Input

Line 1: Two space-separated integers: N andK
Lines 2.. N+1: N integers, one per line, the quality of the milk on dayi appears on the ith line.

Output

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

Sample Input

8 212323231

Sample Output

4

思路：二分长度，判断符合条件不，后缀数组和HASH都行

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))#define LL unsigned long longusing namespace std;const int mm=2e4+9;const int mod=127;LL h[mm],xp[mm],has[mm];int rank[mm],f[mm];int N,K;bool cmp(const int&a,const int&b){  return has[a]<has[b]||(has[a]==has[b]&&a<b);}bool ok(int mid,int k){  for(int i=0;i<N-mid+1;++i)  {    rank[i]=i;    has[i]=h[i]-h[i+mid]*xp[mid];  }  sort(rank,rank+N-mid+1,cmp);  int num=0;  for(int i=0;i<N-mid+1;++i)  {    if(i==0||has[ rank[i] ]!=has[ rank[i-1] ])num=0;    if(++num>=k)return 1;  }  return 0;}int main(){  while(~scanf("%d%d",&N,&K))  {    FOR(i,0,N-1)scanf("%d",&f[i]);    h[N]=0;    for(int i=N-1;i>=0;i--)      h[i]=h[i+1]*mod+f[i];    xp[0]=1;    FOR(i,1,N)xp[i]=xp[i-1]*mod;    int l=1,r=N,mid;    int ans=0;    //if(ok(4,2))puts("+++|");    while(l<=r)    {      mid=(l+r)/2;      if(ok(mid,K))      { ans=max(ans,mid);        l=mid+1;      }      else r=mid-1;    }    printf("%d\n",ans);  }}