POJ 3261 Milk Patterns 可重复k次的最长重复子串

Milk Patterns
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://poj.org/problem?id=3261

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

HINT

题意

让你找到最长的可以重复k次的重复子串

题解：

后缀数组，跑出height数组之后，然后直接二分然后O(n)，check就好了

跑check的时候，只要符合num++就好了，如果大于等于，那就是符合的，然后就好了……

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1005000
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

int s[2*maxn];
int a[maxn];
int n,k;
int sa[maxn], rank[maxn], height[maxn];
int wa[maxn], wb[maxn], wv[maxn], wd[maxn];

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

void build_sa(int *r, int n, int m){          //  倍增算法 r为待匹配数组  n为总长度 m为字符范围
    int i, j, p, *x = wa, *y = wb, *t;
    for(i = 0; i < m; i ++) wd[i] = 0;
    for(i = 0; i < n; i ++) wd[x[i]=r[i]] ++;
    for(i = 1; i < m; i ++) wd[i] += wd[i-1];
    for(i = n-1; i >= 0; i --) sa[-- wd[x[i]]] = i;
    for(j = 1, p = 1; p < n; j *= 2, m = p){
        for(p = 0, i = n-j; i < n; i ++) y[p ++] = i;
        for(i = 0; i < n; i ++) if(sa[i] >= j) y[p ++] = sa[i] - j;
        for(i = 0; i < n; i ++) wv[i] = x[y[i]];
        for(i = 0; i < m; i ++) wd[i] = 0;
        for(i = 0; i < n; i ++) wd[wv[i]] ++;
        for(i = 1; i < m; i ++) wd[i] += wd[i-1];
        for(i = n-1; i >= 0; i --) sa[-- wd[wv[i]]] = y[i];
        for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i ++){
            x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p - 1: p ++;
        }
    }
}

void calHeight(int *r, int n){           //  求height数组。
    int i, j, k = 0;
    for(i = 1; i <= n; i ++) rank[sa[i]] = i;
    for(i = 0; i < n; height[rank[i ++]] = k){
        for(k ? k -- : 0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k ++);
    }
}
int check(int mid)
{
    int num=0;
    for(int i=0;i<n;i++)
    {
        if(height[i]>=mid)
            num++;
        else
            num=1;
        if(num>=k)
            return 1;
    }
    return 0;
}
int main()
{
    n=read(),k=read();
    for(int i=0;i<n;i++)
        a[i]=read(),a[i]++;
    a[n++]=0;
    build_sa(a,n,1001000);
    calHeight(a,n);
    int l=0,r=n;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(check(mid))
        {
            l=mid+1;
        }
        else
            r=mid-1;
    }
    cout<<r<<endl;
}