poj3261 Milk Patterns

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 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

和poj1743【见这里】一样，跑完后缀数组用二分答案分块，这次要验证的是是否有一块的长度不小于要求的重复次数。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[20010],sa[20010],rank[20010],height[20010],t1[20010],t2[20010],cnt[20010],ord[20010],m,n,k;
void init()
{
    int i;
    scanf("%d%d",&n,&k);
    for (i=1;i<=n;i++)
      scanf("%d",&a[i]),ord[i]=a[i];
    sort(ord+1,ord+n+1);
    m=unique(ord+1,ord+n+1)-ord-1;
    for (i=1;i<=n;i++)
      a[i]=lower_bound(ord+1,ord+m+1,a[i])-ord;
}
void make()
{
    int i,j,k,p,*x=t1,*y=t2;
    for (i=1;i<=n;i++)
      cnt[x[i]=a[i]]++;
    for (i=2;i<=m;i++)
      cnt[i]+=cnt[i-1];
    for (i=n;i;i--)
      sa[cnt[x[i]]--]=i;
    for (k=1;k<=n;k<<=1)
    {
        p=0;
        for (i=n-k+1;i<=n;i++)
          y[++p]=i;
        for (i=1;i<=n;i++)
          if (sa[i]-k>=1)
            y[++p]=sa[i]-k;
        for (i=1;i<=m;i++)
          cnt[i]=0;
        for (i=1;i<=n;i++)
          cnt[x[y[i]]]++;
        for (i=2;i<=m;i++)
          cnt[i]+=cnt[i-1];
        for (i=n;i;i--)
          sa[cnt[x[y[i]]]--]=y[i];
        swap(x,y);
        x[sa[1]]=p=1;
        for (i=2;i<=n;i++)
          if (y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k])
            x[sa[i]]=p;
          else x[sa[i]]=++p;
        if ((m=p)>=n) break;
    }
    for (i=1;i<=n;i++)
      rank[sa[i]]=i;
    for (i=1,k=0;i<=n;i++)
    {
        if (k) k--;
        if (rank[i]==1)
        {
            k=0;
            continue;
        }
        while (a[i+k]==a[sa[rank[i]-1]+k]) k++;
        height[rank[i]]=k;
    }
}
bool ok(int x)
{
    int i,l=1;
    for (i=2;i<=n;i++)
      if (height[i]<x)
      {
        if (i-l>=k) return 1;
        l=i;
      }
    return n-l+1>=k;
}
void solve()
{
    int l=1,r=n,mid;
    while (l<r)
    {
        mid=(l+r+1)/2;
        if (ok(mid)) l=mid;
        else r=mid-1;
    }
    printf("%d\n",l);
}
int main()
{
    init();
    make();
    solve();
}