poj3261Milk Patterns 二分+后缀数组

最新推荐文章于 2021-07-19 09:45:07 发布

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最新推荐文章于 2021-07-19 09:45:07 发布

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分类专栏：后缀数组文章标签：后缀数组二分

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本文介绍了一种通过二分查找和后缀数组技术来解决寻找字符串中最长重复子串的问题，该方法允许子串间相互重叠，并确保了高效的算法性能。

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Milk Patterns

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 9665		Accepted: 4333
Case Time Limit: 2000MS

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

Sample Input

Sample Output

Source

USACO 2006 December Gold

Language:

题意：给定一个字符串，求至少出现k次的最长重复子串，这k个子串可以重叠。

思路：二分答案，利用height数组，将后缀分成若干组，判断是否存在一个组的后缀个数不小于k，如果存在那么存在k个相同的子串满足条件。时间复杂度为O（nlogn），详见代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=20000+100;
const int M=1000010;
int n,K;
int a[MAXN],sa[MAXN],height[MAXN],c[M],rank[MAXN],t1[MAXN],t2[MAXN];
int cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b] && r[a+l]==r[b+l];
}
void build_sa(int m)
{
    int i,k,p=0;
    int *x=t1,*y=t2;
    for(i=0;i<m;i++) c[i]=0;
    for(i=0;i<n;i++) c[x[i]=a[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(k=1;p<n;k<<=1,m=p)
    {
        for(p=0,i=n-k;i<n;i++) y[p++]=i;
        for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
        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);
        for(p=1,x[sa[0]]=0,i=1;i<n;i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],k)? p-1:p++;
    }
}
bool ok(int n,int m)
{
    int cnt=1;
    for(int i=1;i<=n;i++)
    {
        if(height[i]>=m)
        {
            cnt++;
            if(cnt>=K)
                return 1;
        }
        else
            cnt=1;
    }
    return 0;
}
void calheight(int n)
{
    int k=0;
    for(int i=1;i<=n;i++) rank[sa[i]]=i;
    for(int i=0;i<n;i++)
    {
        if(k) k--;
        int j=sa[rank[i]-1];
        while(a[i+k]==a[j+k]) k++;
        height[rank[i]]=k;
    }
}
int main()
{
    //freopen("text.txt","r",stdin);
    while(~scanf("%d%d",&n,&K))
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]),a[i]++;
        a[n++]=0;
        build_sa(1000010);
        calheight(n-1);
        int l=1,r=n-1,ans=0;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(ok(n-1,mid))
                ans=mid,l=mid+1;
            else
                r=mid-1;
        }
        printf("%d\n",ans);
    }
    return 0;
}