[USACO2006 DEC] Milk Patterns

[题目链接]

        https://www.lydsy.com/JudgeOnline/problem.php?id=1717

[算法]

       首先二分答案 ， 然后将后缀分组即可

       详见2009国家集训队论文集之 : 《后缀数组——处理字符串的有利工具》

       时间复杂度 : O(NlogN)

[代码]

       

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 1500000
 
int n , k;
int height[MAXN] , cnt[MAXN] , rk[MAXN] , sa[MAXN] , a[MAXN] , x[MAXN] , y[MAXN];
 
template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); } 
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f; 
} 
inline void build_sa()
{
    memset(cnt , 0 , sizeof(cnt));
    for (int i = 1; i <= n; i++) ++cnt[a[i]];
    for (int i = 1; i <= 1000010; i++) cnt[i] += cnt[i - 1];
    for (int i = 1000010; i >= 1; i--) sa[cnt[a[i]]--] = i;
    rk[sa[1]] = 1;
    for (int i = 2; i <= n; i++) rk[sa[i]] = rk[sa[i - 1]] + (a[sa[i - 1]] != a[sa[i]]);
    for (int k = 1; rk[sa[n]] != n; k <<= 1)
    {
        for (int i = 1; i <= n; i++)
            x[i] = rk[i] , y[i] = (i + k <= n) ? rk[i + k] : 0;
        for (int i = 0; i <= n; i++) cnt[i] = 0;
        for (int i = 1; i <= n; i++) ++cnt[y[i]];
        for (int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
        for (int i = n; i >= 1; i--) rk[cnt[y[i]]--] = i;
        for (int i = 1; i <= n; i++) cnt[i] = 0;
        for (int i = 1; i <= n; i++) ++cnt[x[i]];
        for (int i = 0; i <= n; i++) cnt[i] += cnt[i - 1];
        for (int i = n; i >= 1; i--) sa[cnt[x[rk[i]]]--] = rk[i];
        rk[sa[1]] = 1;
        for (int i = 2; i <= n; i++) rk[sa[i]] = rk[sa[i - 1]] + (x[sa[i]] != x[sa[i - 1]] || y[sa[i]] != y[sa[i - 1]]);
    }
}
inline void get_height()
{
    int k = 0;
    for (int i = 1; i <= n; i++)
    {
        if (k) --k;
        int j = sa[rk[i] - 1];
        while (a[i + k] == a[j + k]) ++k;
        height[rk[i]] = k;  
    }   
} 
inline bool check(int mid)
{
    int cnt = 1;
    for (int i = 2; i <= n; i++)
    {
        if (height[i] >= mid) 
        {
            ++cnt;
            if (cnt >= k) return true;
        } else cnt = 1;
    }
    return false;
}
  
int main()
{
     
    read(n); read(k);
    for (int i = 1; i <= n; i++) read(a[i]);
    build_sa();
    get_height();
    int l = 1 , r = n , ans = 0;
    while (l <= r)
    {
        int mid = (l + r) >> 1;
        if (check(mid))
        {
            ans = mid;
            l = mid + 1;
        } else r = mid - 1;
    }
    for (int i = 1; i <= n; i++)
    {
        ++cnt[a[i]];
        if (cnt[a[i]] >= k) chkmax(ans , 1);
    }
    printf("%d\n" , ans);
     
    return 0;
}