poj3261Milk Patterns 后缀数组

题目地址：http://poj.org/problem?id=3261

题目：

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

思路：直接上后缀数组模板，跑个height数组。不过此时要注意两点：
　　1.每个数的范围在0~1e6，所以要先离散化。不过此题数据较水，你可以不需要离散化，直接把字符最大值设为1e5就可以a了。不过直接开个1e6的数组应该也没事吧（没试过，待证实）
　　2.这个数可能是0，所以要么把每个读取的数先+1，或者把ss[n]设为-1；
　　然后二分答案，对height数组分组看符不符合答案。

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

const int N = 200000+9;
int wa[N], wb[N], ws[N], wv[N];
int rank[N], height[N];
int sa[N],ss[N],n,k;
bool cmp(int r[], int a, int b, int l) {
    return r[a] == r[b] && r[a+l] == r[b+l];
}

void da(int r[], int sa[], int n, int m) {
    int i, j, p, *x = wa, *y = wb;
    for (i = 0; i < m; ++i) ws[i] = 0;
    for (i = 0; i < n; ++i) ws[x[i]=r[i]]++;
    for (i = 1; i < m; ++i) ws[i] += ws[i-1];
    for (i = n-1; i >= 0; --i) sa[--ws[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) ws[i] = 0;
        for (i = 0; i < n; ++i) ws[wv[i]]++;
        for (i = 1; i < m; ++i) ws[i] += ws[i-1];
        for (i = n-1; i >= 0; --i) sa[--ws[wv[i]]] = y[i];
        for (swap(x, y), 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 sa[], int n) {
    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 len)
{
    int i=2,cnt=0;
    while(1)
    {
        while(i<=n && height[i]>=len)
            cnt++,i++;
        if(cnt+1>=k)return 1;
        if(i>=n)return 0;
        while(i <=n &&height[i]<len)
            i++;
        cnt=0;
    }
}

int main()
{
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
        scanf("%d",&ss[i]),ss[i]++;
    ss[n]=0;
    da(ss,sa,n+1,N);
    calheight(ss,sa,n);
    int l=1,r=n,ans=1,mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(check(mid))
            l=mid+1,ans=mid;
        else
            r=mid-1;
    }
    printf("%d\n",ans);
    return 0;
}

 

转载于:https://www.cnblogs.com/weeping/p/5751422.html