pku3261 Milk Patterns 后缀数组/可重叠的 k 次最长重复子串

Milk Patterns

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 16696 Accepted: 7371Case 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

8 212323231

Sample Output

4

Source

USACO 2006 December Gold

先二分答案,然后将后缀分成若干组。不
同的是,这里要判断的是有没有一个组的后缀个数不小于 k。如果有,那么存在
k 个相同的子串满足条件,否则不存在

//#include<bits/stdc++.h>#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<set>#include<ctime>#include<vector>#include<cmath>#include<algorithm>using namespace std;typedef long long ll;const int MAXN = 20005;/**suffix array*倍增算法  O(n*logn)*待排序数组长度为n,放在0~n-1中，在最后面补一个0*build_sa( ,n+1, );//注意是n+1;*getHeight(,n);*例如：*n   = 8;*num[]   = { 1, 1, 2, 1, 1, 1, 1, 2, $ };注意num最后一位为0，其他大于0*rank[]  = { 4, 6, 8, 1, 2, 3, 5, 7, 0 };rank[0~n-1]为有效值，rank[n]必定为0无效值*sa[]    = { 8, 3, 4, 5, 0, 6, 1, 7, 2 };sa[1~n]为有效值，sa[0]必定为n是无效值*height[]= { 0, 0, 3, 2, 3, 1, 2, 0, 1 };height[2~n]为有效值**/int sa[MAXN];//SA数组，表示将S的n个后缀从小到大排序后把排好序的             //的后缀的开头位置顺次放入SA中int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值int rank1[MAXN],height[MAXN];//待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,//除s[n-1]外的所有s[i]都大于0，r[n-1]=0//函数结束以后结果放在sa数组中void build_sa(int s[],int n,int m){    int i,j,p,*x=t1,*y=t2;    //第一轮基数排序，如果s的最大值很大，可改为快速排序    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;        //直接利用sa数组排序第二关键字        for(i=n-j;i<n;i++)y[p++]=i;//后面的j个数第二关键字为空的最小        for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;        //这样数组y保存的就是按照第二关键字排序的结果        //基数排序第一关键字        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];        //根据sa和x数组计算新的x数组        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;//下次基数排序的最大值    }}void getHeight(int s[],int n){    int i,j,k=0;    for(i=0;i<=n;i++)rank1[sa[i]]=i;    for(i=0;i<n;i++)    {        if(k)k--;        j=sa[rank1[i]-1];        while(s[i+k]==s[j+k])k++;        height[rank1[i]]=k;    }}int str[MAXN];int s[MAXN];int n,k;bool check(int len){    int l,r;    l = 2,r=2;    while(l<=n&&r<=n)    {        if(height[l]<len)        {            l++;            continue;        }        if(height[l]>=len)        {            r=l;            while(height[r]>=len && r<=n) r++;        }        if(r-l+1>=k)        {            return true;        }        l=r;    }//    cout <<"len : "<<len<<endl;    return false;}int main(){//    freopen("data.txt","r",stdin);    while(scanf("%d %d",&n,&k)==2)    {        for(int i=0;i<n;i++)            scanf("%d",&s[i]);        s[n]=0;        build_sa(s,n+1,250);        getHeight(s,n);        int ans=0;        int l = 1;        int r = n;        while(l<=r)        {            int mid=(l+r)>>1;            if(check(mid))                ans=mid,l=mid+1;            else                r=mid-1;        }        printf("%d\n",ans);    }    return 0;}