POJ3261---Milk Patterns（后缀数组+二分）

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

求最长可重叠重复k次的串
建立后缀数组，然后二分答案，对后缀分组，看当前长度下，是否存在某一组，后缀数目大于等于k个

/*************************************************************************
    > File Name: POJ3261.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年03月31日 星期二 16时43分03秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

class SuffixArray
{
    public:
        static const int N = 22000;
        int init[N];
        int X[N];
        int Y[N];
        int Rank[N];
        int sa[N];
        int height[N];
        int buc[N];
        int size;

        void clear()
        {
            size = 0;
        }

        void insert(int n)
        {
            init[size++] = n;
        }

        bool cmp(int *r, int a, int b, int l)
        {
            return (r[a] == r[b] && r[a + l] == r[b + l]);
        }

        void getsa(int m)
        {
            init[size] = 0;
            int l, p, *x = X, *y = Y, n = size + 1;
            for (int i = 0; i < m; ++i)
            {
                buc[i] = 0;
            }
            for (int i = 0; i < n; ++i)
            {
                ++buc[x[i] = init[i]];
            }
            for (int i = 1; i < m; ++i)
            {
                buc[i] += buc[i - 1];
            }
            for (int i = n - 1; i >= 0; --i)
            {
                sa[--buc[x[i]]] = i;
            }
            for (l = 1, p = 1; l <= n && p < n; m = p, l *= 2)
            {
                p = 0;
                for (int i = n - l; i < n; ++i)
                {
                    y[p++] = i;
                }
                for (int i = 0; i < n; ++i)
                {
                    if (sa[i] >= l)
                    {
                        y[p++] = sa[i] - l;
                    }
                }
                for (int i = 0; i < m; ++i)
                {
                    buc[i] = 0;
                }
                for (int i = 0; i < n; ++i)
                {
                    ++buc[x[y[i]]];
                }
                for (int i = 1; i < m; ++i)
                {
                    buc[i] += buc[i - 1];
                }
                for (int i = n - 1; i >= 0; --i)
                {
                    sa[--buc[x[y[i]]]] = y[i];
                }
                int i;
                for (swap(x, y), x[sa[0]] = 0, p = 1, i = 1; i < n; ++i)
                { 
                    x[sa[i]] = cmp(y, sa[i - 1], sa[i], l) ? p - 1 : p++; 
                }
            }
        }

        void getheight()
        {
            int h = 0, n = size;
            for (int i = 0; i <= n; ++i)
            {
                Rank[sa[i]] = i;
            }
            height[0] = 0;
            for (int i = 0; i < n; ++i)
            {
                if (h > 0)
                {
                    --h;
                }
                int j =sa[Rank[i] - 1];
                for (; i + h < n && j + h < n && init[i + h] == init[j + h]; ++h);
                height[Rank[i] - 1] = h;
            }
        }

        bool check(int k, int l)
        {
            int cnt = 1;
            for (int i = 0; i < size; ++i)
            {
                if (height[i] >= l)
                {
                    ++cnt;
                    if (cnt >= k)
                    {
                        return 1;
                    }
                }
                else
                {
                    if (cnt >= k)
                    {
                        return 1;
                    }
                    cnt = 1;
                }
            }
            return 0;
        }

        void solve(int k)
        {
            int l = 1, r = size;
            int mid;
            int ans = 0;
            while (l <= r)
            {
                mid = (l + r) >> 1;
                if (check(k, mid))
                {
                    ans = mid;
                    l = mid + 1;
                }
                else
                {
                    r = mid - 1;
                }
            }
            printf("%d\n", ans);
        }
}SA;

int main()
{
    int n, k;
    while (~scanf("%d%d", &n, &k))
    {
        int maxs = -1, val;
        SA.clear();
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &val);
            SA.insert(val);
            maxs = max(maxs, val);
        }
        SA.getsa(maxs + 1);
        SA.getheight();
        SA.solve(k);
    }
    return 0;
}