poj 3261 Milk Patterns 后缀数组 可重叠的k次最长重复子串

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.

Input
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.
Output
Line 1: One integer, the length of the longest pattern which occurs at least K times
Sample Input
8 2
1
2
3
2
3
2
3
1
Sample Output
4

数据1e6 上板子

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 1e6+100;int t1[MAXN],t2[MAXN],c[MAXN];bool cmp(int *r,int a,int b,int l){    return r[a]==r[b]&&r[a+l]==r[b+l];}void da(int str[],int sa[],int ra[],int height[],int n,int m){    n++;    int i,j,p,*x=t1,*y=t2;    for(i=0;i<m;i++) c[i]=0;    for(i=0;i<n;i++) c[x[i]=str[i]]++;    for(i=1;i<m;i++) c[i]+=c[i-1];    for(i=n-1;i>=0;i--) sa[--c[x[i]]]=i;    for(j=1;j<=n;j<<=1)    {        p=0;        for(i=n-j;i<n;i++) y[p++]=i;        for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;        for(i=0;i<m;i++) c[i]=0;        for(i=0;i<n;i++) c[x[y[i]]]++;        for(i=1;i<m;i++) c[i]+=c[i-1];        for(i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];        swap(x,y);        p=1;x[sa[0]]=0;        for(i=1;i<n;i++)            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;        if(p>=n) break;        m=p;    }    int k=0;    n--;    for(i=0;i<=n;i++) ra[sa[i]]=i;    for(i=0;i<n;i++)     {        if(k) k--;        j=sa[ra[i]-1];        while(str[i+k]==str[j+k])k++;        height[ra[i]]=k;    }}int ra[MAXN],height[MAXN];int sa[MAXN],num[MAXN];int n,k;int check(int x){    int i=1;    while(i<n)    {        int j=i+1,cnt=0;        while(j<=n&&height[j]>=x)        {            cnt++;            j++;        }        if(cnt>=k-1) return 1;        i=j;    }    return 0;}int main(){    while(scanf("%d%d",&n,&k)!=EOF)    {        // int maxn=0;        for(int i=0;i<n;i++)        {            scanf("%d",&num[i]);            // maxn=max(num[i],maxn);        }        num[n]=0;        da(num,sa,ra,height,n,MAXN);        int l=0,r=n;        while(l<=r)        {            int mid=(l+r)>>1;            if(check(mid))                l=mid+1;            else r=mid-1;        }        printf("%d\n",l-1 );    }}