POJ 3261 Milk Patterns （后缀数组，求可重叠的k次最长重复子串）

青旅半醒 2022-03-27 10:25 129阅读 0赞

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      Milk Patterns 
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        <td><strong>Time Limit:</strong>&nbsp;5000MS</td> 
        <td>&nbsp;</td> 
        <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
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        <td><strong>Total Submissions:</strong>&nbsp;7586</td> 
        <td>&nbsp;</td> 
        <td><strong>Accepted:</strong>&nbsp;3448</td> 
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        <td colspan="3" align="center"><strong>Case Time Limit:</strong>&nbsp;2000MS</td> 
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    </div> <p>Description</p> 
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     <p>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.</p> 
     <p>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&nbsp;<em>N</em>&nbsp;(1 ≤&nbsp;<em>N</em>&nbsp;≤ 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 ≤&nbsp;<em>K</em>&nbsp;≤&nbsp;<em>N</em>) times. This may include overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example.</p> 
     <p>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&nbsp;<em>K</em>&nbsp;times.</p> 
    </div> <p>Input</p> 
    <div>
      Line 1: Two space-separated integers:&nbsp; 
     <em>N</em>&nbsp;and&nbsp; 
     <em>K</em>&nbsp; 
     <br>Lines 2.. 
     <em>N</em>+1:&nbsp; 
     <em>N</em>&nbsp;integers, one per line, the quality of the milk on day&nbsp; 
     <em>i</em>&nbsp;appears on the&nbsp; 
     <em>i</em>th line. 
    </div> <p>Output</p> 
    <div>
      Line 1: One integer, the length of the longest pattern which occurs at least&nbsp; 
     <em>K</em>&nbsp;times 
    </div> <p>Sample Input</p> <pre>8 2
1
2
3
2
3
2
3
1</pre> <p>Sample Output</p> <pre>4</pre> <p>Source</p> 
    <div> 
     <a href="http://poj.org/searchproblem?field=source&amp;key=USACO+2006+December+Gold" rel="nofollow">USACO 2006 December Gold</a> 
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      &nbsp; 
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      &nbsp; 
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      &nbsp; 
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**后缀数组 的入门题。**

**先计算好SA和height数组。**

**然后二分答案。二分的时候，对height进行分组，看存不存在一组的大小大于等于k**

/*
     * POJ 3261 Milk Patterns
     * 可重叠的k次最长重复子串
     * 利用后缀数组计算S和height数组
     * 然后二分，进行分组
     */
    
    #include <iostream>
    #include <string.h>
    #include <stdio.h>
    #include <algorithm>
    using namespace std;
    const int MAXN=20010;
    
    int sa[MAXN];
    int t1[MAXN],t2[MAXN],c[MAXN];
    int rank[MAXN],height[MAXN];
    void build_sa(int s[],int n,int m)
    {
        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]=s[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]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
            if(p>=n)break;
            m=p;
        }
    }
    void getHeight(int s[],int n)
    {
        int i,j,k=0;
        for(i=0;i<=n;i++)rank[sa[i]]=i;
        for(i=0;i<n;i++)
        {
            if(k)k--;
            j=sa[rank[i]-1];
            while(s[i+k]==s[j+k])k++;
            height[rank[i]]=k;
        }
    }
    
    bool check(int n,int k,int t)
    {
        int num=1;
        for(int i=2;i<=n;i++)
        {
            if(height[i]>=t)
            {
                num++;
                if(num>=k)return true;
            }
            else num=1;
        }
        return false;
    }
    int s[MAXN];
    int main()
    {
        //freopen("in.txt","r",stdin);
        //freopen("out.txt","w",stdout);
        int n,k;
        while(scanf("%d%d",&n,&k)==2)
        {
            int Max=0;
            for(int i=0;i<n;i++)
            {
                scanf("%d",&s[i]);
                Max=max(Max,s[i]);
            }
            s[n]=0;
            build_sa(s,n+1,Max+1);
            getHeight(s,n);
            int l=0,r=n;
            int ans=0;
            while(l<=r)
            {
                int mid=(l+r)/2;
                if(check(n,k,mid))
                {
                    ans=mid;
                    l=mid+1;
                }
                else r=mid-1;
            }
            printf("%d\n",ans);
        }
        return 0;
    }