ase erl href lin == noticed per ember 數組

題目：

Milk Patterns

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 16482		Accepted: 7283
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 翻譯：（from bzoj-1717） 農夫John發現他的奶牛產奶的質量一直在變動。經過細致的調查，他發現：雖然他不能預見明天 產奶的質量，但連續的若幹天的質量有很多重疊。我們稱之為一個“模式”。 John的牛奶按質量可以被賦予一個0到1000000之間的數。並且John記錄了N(1<=N<=20000)天的 牛奶質量值。他想知道最長的出現了至少K(2<=K<=N)次的模式的長度。 比如1 2 3 2 3 2 3 1 中 2 3 2 3出現了兩次。當K=2時，這個長度為4。 Input * Line 1: 兩個整數 N,K。 * Lines 2..N+1: 每行一個整數表示當天的質量值。 Output * Line 1: 一個整數：N天中最長的出現了至少K次的模式的長度 分析： 後綴數組的height[i]儲存的是字典序排名為i的串和字典序排名位i-1的串的最長公共前綴。 那麽我們二分答案d，檢測是否有連續k-1個height元素>d即可。 代碼：

#pragma warning(disable:4786)
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<cmath>
#include<string>
#include<sstream>
#define LL long long
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define mem(a,x) memset(a,x,sizeof(a))
#define lson l,m,x<<1
#define rson m+1,r,x<<1|1
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3e4 + 5;
const int maxs = 30;
int s[maxn];
int sa[maxn], t[maxn], t2[maxn], c[maxn], n , m;
int d[maxn][maxs];
int height[maxn], rank[maxn];
void build_sa(int m)
{
	int i, *x = t, *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 (int k = 1; k <= n; k <<= 1){
		int p = 0;
		for (i = n - 1; i >= n-k; i--)
			y[p++] = i;
		for (i = 0; i < n; i++){
			if (sa[i] >= k)
				y[p++] = sa[i] - k;
		}
		for (i = 0; i < m; i++)	c[i] = 0;
		for (i = 0; i < n; i++)
			c[x[y[i]]]++;
		for (i = 0; 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] + k] == y[sa[i] + k] ? p - 1 : p++;
		}
		if (p >= n)	break;
		m = p;
	}
}
void getHeight()
{
	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;
	}
}
void RMQ_init()
{
    for(int i = 0 ; i<n; i++)       d[i][0] = sa[i];
    for(int j = 1 ; (1<<j) - 1 <=n ; j++){
        for(int i = 0 ; i + (1<<j) - 1 < n ; i++){
            d[i][j] = min(d[i][j-1] , d[i + (1<< (j - 1))][ j -1 ]);
        }
    }
}
int RMQ(int L , int R)
{
    int k = 0;
    while((1<<(k + 1)) <= R - L + 1)    k++;
    return min(d[L][k] , d[R - (1 << k) + 1][k]);
}
int k;
int check(int d){
    int l=0;
    for (int i=1;i<n;i++){
        if (height[i]>=d){
            if (l==0) l=2;
            else l++;
        }else{
        l=0;
        }
        if (l>=k) return 1;
    }
    return 0;
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF){
        int maxm=0,p;
        for (int i=1;i<=n;i++){
            scanf("%d",&p);
            s[i-1]=p;
            if (p>maxm) maxm=p;
        }
        build_sa(maxm+1);
        getHeight();
        int l=0,r=n;
        int m=(l+r)/2;
        int zans=0;
        while(l<=r){
            m=(l+r)/2;
            int ans=check(m);
            if (ans) l=m+1,zans=m;
            else
            r=m-1;
        }
        printf("%d\n",zans);
    }
}

POJ - 3261 Milk Patterns 後綴數組