poj3261 Milk Patterns

Milk Patterns

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 17196		Accepted: 7612
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

8 2
1
2
3
2
3
2
3
1

Sample Output

4

Source

USACO 2006 December Gold

题目大意：求一个最长的重复出现了至少k次的连续子串，它们可以重叠.

分析：类似poj1743的方法，还是二分+分组.在check函数上改动一下就好了.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 30010;
int n,s[maxn],ans,fir[maxn],sec[maxn],pos[maxn],sa[maxn],rk[maxn],tong[maxn],ht[maxn];
int sett[maxn],a[maxn],cnt,K;

void solve()
{
    int len = n;
    memset(rk,0,sizeof(rk));
    memset(sa,0,sizeof(sa));
    memset(ht,0,sizeof(ht));
    memset(fir,0,sizeof(fir));
    memset(sec,0,sizeof(sec));
    memset(pos,0,sizeof(pos));
    memset(tong,0,sizeof(tong));
    copy(s + 1,s + len + 1,sett + 1);
    sort(sett + 1,sett + 1 + len);
    cnt = unique(sett + 1,sett + 1 + len) - sett - 1;
    for (int i = 1; i <= len; i++)
        a[i] = lower_bound(sett + 1,sett + 1 + cnt,s[i]) - sett;
    for (int i = 1; i <= len; i++)
        tong[a[i]]++;
    for (int i = 1; i <= len; i++)
        tong[i] += tong[i - 1];
    for (int i = 1; i <= len; i++)
        rk[i] = tong[a[i] - 1] + 1;
    for (int t = 1; t <= len; t *= 2)
    {
        for (int i = 1; i <= len; i++)
            fir[i] = rk[i];
        for (int i = 1; i <= len; i++)
        {
            if (i + t > len)
                sec[i] = 0;
            else
                sec[i] = rk[i + t];
        }
        fill(tong,tong + 1 + len,0);
        for (int i = 1; i <= len; i++)
            tong[sec[i]]++;
        for (int i = 1; i <= len; i++)
            tong[i] += tong[i - 1];
        for (int i = 1; i <= len; i++)
            pos[len - --tong[sec[i]]] = i;
        fill(tong,tong + 1 + len,0);
        for (int i = 1; i <= len; i++)
            tong[fir[i]]++;
        for (int i = 1; i <= len; i++)
            tong[i] += tong[i - 1];
        for (int i = 1; i <= len; i++)
        {
            int temp = pos[i];
            sa[tong[fir[temp]]--] = temp;
        }
        bool flag = true;
        int last = 0;
        for (int i = 1; i <= len; i++)
        {
            int temp = sa[i];
            if (!last)
                rk[temp] = 1;
            else if (fir[temp] == fir[last] && sec[temp] == sec[last])
            {
                rk[temp] = rk[last];
                flag = false;
            }
            else
                rk[temp] = rk[last] + 1;
            last = temp;
        }
        if (flag)
            break;
    }
    int k = 0;
    for (int i = 1; i <= len; i++)
    {
        if (rk[i] == 1)
            k = 0;
        else
        {
            if (k)
                k--;
            int j = sa[rk[i] - 1];
            while (i + k <= len && j + k <= len && a[i + k] == a[j + k])
                k++;
        }
        ht[rk[i]] = k;
    }
}

bool check(int x)
{
    int pos = 1;
    for (int i = 2; i <= n; i++)
    {
        if (ht[i] >= x)  //代表重复子串的长度
        {
            if (i - pos + 1 >= K)
                return true;
            continue;
        }
        pos = i;
    }
    return false;
}

int main()
{
    scanf("%d%d",&n,&K);
    for (int i = 1; i <= n; i++)
        scanf("%d",&s[i]);
    solve();
    int l = 1,r = n;
    while (l <= r)
    {
        int mid = (l + r) >> 1;
        if (check(mid))
        {
            ans = mid;
            l = mid + 1;
        }
        else
            r = mid - 1;
    }
    printf("%d
",ans);

    return 0;
}