poj3261Milk Patterns【可重叠最长重复子串次数大于k】

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 12892 Accepted: 5740Case 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 overN (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 leastK times.

Input

Line 1: Two space-separated integers: N andK
Lines 2..N+1: N integers, one per line, the quality of the milk on dayi appears on theith line.

Output

Line 1: One integer, the length of the longest pattern which occurs at leastK times

Sample Input

8 212323231

Sample Output

4

Source

USACO 2006 December Gold

和1743我的上一篇差不多，不可重叠改成可重叠，那么就不用劳心费神的要求两个位置必须大于重复单元长度了，只要是组内的个数大于已知值就可以了

/*************poj32612016.2.221236K63MSG++2399B*************/#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;const int MAXN=20010;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 rank[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++)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;    }}int s[MAXN];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 main(){    //freopen("cin.txt","r",stdin);    //freopen("out.txt","w",stdout);    int n,k;    while(~scanf("%d%d",&n,&k))    {        int maxn=0;        for(int i=0;i<n;i++){            scanf("%d",&s[i]);            if(maxn<s[i]) maxn=s[i];        }        s[n]=0;//必须是0！        //printf("maxn=%d ",maxn);        build_sa(s,n+1,maxn+2);        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;}