BZOJ 1717: [Usaco2006 Dec]Milk Patterns 产奶的模式

1717: [Usaco2006 Dec]Milk Patterns 产奶的模式

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 1066  Solved: 582
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Description

农夫John发现他的奶牛产奶的质量一直在变动。经过细致的调查，他发现：虽然他不能预见明天产奶的质量，但连续的若干天的质量有很多重叠。我们称之为一个“模式”。 John的牛奶按质量可以被赋予一个0到1000000之间的数。并且John记录了N(1<=N<=20000)天的牛奶质量值。他想知道最长的出现了至少K(2<=K<=N)次的模式的长度。比如1 2 3 2 3 2 3 1 中 2 3 2 3出现了两次。当K=2时，这个长度为4。

Input

* Line 1: 两个整数 N,K。

* Lines 2..N+1: 每行一个整数表示当天的质量值。

Output

* Line 1: 一个整数：N天中最长的出现了至少K次的模式的长度

Sample Input

8 2
1
2
3
2
3
2
3
1

Sample Output

4

HINT

 

Source

Gold

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求出后缀数组,二分答案,在height数组里找出最大的出现次数进行判断

 

#include <bits/stdc++.h>

inline char Char(void)
{
    static const int siz = 1 << 20;
    
    static char buf[siz];
    static char *hd = buf + siz;
    static char *tl = buf + siz;
    
    if (hd == tl)
        fread(hd = buf, 1, siz, stdin);
    
    return *hd++;
}

inline int Int(void)
{
    int ret = 0, neg = 0, c = Char();
    
    for (; c < 48; c = Char())
        if (c == '-')neg ^= true;
    
    for (; c > 47; c = Char())
        ret = ret * 10 + c - '0';
    
    return neg ? -ret : ret;
}

const int siz = 20005;

int n, m, s[siz];

int map[siz], tot;

int sa[siz], rk[siz], ht[siz];

inline void prework_sa(void)
{
    static int ta[siz];
    static int wa[siz], wb[siz];
    static int ca[siz], cb[siz];
    
    for (int i = 1; i <= n; ++i)
        ++ca[s[i]];
    
    for (int i = 1; i <= n; ++i)
        ca[i] += ca[i - 1];
    
    for (int i = n; i >= 1; --i)
        sa[ca[s[i]]--] = i;
    
    rk[sa[1]] = 1;
    
    for (int i = 2; i <= n; ++i)
    {
        rk[sa[i]] = rk[sa[i - 1]];
        if (s[sa[i]] != s[sa[i - 1]])
            ++rk[sa[i]];
    }
    
    for (int l = 1; rk[sa[n]] < n; l <<= 1)
    {
        for (int i = 1; i <= n; ++i)
            ca[i] = cb[i] = 0;
        
        for (int i = 1; i <= n; ++i)
        {
            ++ca[wa[i] = rk[i]];
            ++cb[wb[i] = i + l <= n ? rk[i + l] : 0];
        }
        
        for (int i = 1; i <= n; ++i)
        {
            ca[i] += ca[i - 1];
            cb[i] += cb[i - 1];
        }
        
        for (int i = n; i >= 1; --i)
            ta[cb[wb[i]]--] = i;
        
        for (int i = n; i >= 1; --i)
            sa[ca[wa[ta[i]]]--] = ta[i];
        
        rk[sa[1]] = 1;
        
        for (int i = 2; i <= n; ++i)
        {
            rk[sa[i]] = rk[sa[i - 1]];
            if (wa[sa[i]] != wa[sa[i - 1]]
            ||  wb[sa[i]] != wb[sa[i - 1]])
                ++rk[sa[i]];
        }
    }
    
    for (int i = 1, j = 0; i <= n; ++i)
    {
        if (--j < 0)j = 0;
        while (s[i + j] == s[sa[rk[i] - 1] + j])++j;
        ht[rk[i]] = j;
    }
}

inline int count(int len)
{
    int ret = 0, cnt = 1;
    
    for (int i = 2; i <= n; ++i)
        if (ht[i] >= len)
            ++cnt;
        else
        {
            if (ret < cnt)
                ret = cnt;
            cnt = 1;
        }
    
    if (ret < cnt)
        ret = cnt;
    
    return ret;
}

signed main(void)
{
    n = Int();
    m = Int();
    
    for (int i = 1; i <= n; ++i)
        s[i] = map[++tot] = Int();
    
    std::sort(map + 1, map + tot + 1);
    
    tot = std::unique(map + 1, map + tot + 1) - map;
    
    for (int i = 1; i <= n; ++i)
        s[i] = std::lower_bound(map + 1, map + tot, s[i]) - map;
        
    prework_sa();
    
    int lt = 0, rt = n, mid, ans = 0;
    
    while (lt <= rt)
    {
        mid = (lt + rt) >> 1;
        
        if (count(mid) >= m)
            lt = mid + 1, ans = mid;
        else
            rt = mid - 1;
    }
    
    printf("%d\n", ans);
}

 

@Author: YouSiki