Resource limit
Time limit: 100ms Memory limit: 256.0MB
Problem description
Given a number string of length n, find the length of the longest repeated substring that appears at least k times. These k substrings can overlap. It is guaranteed that the substring appears at least k times.
Input format The
first line: two integers n, k; the
second line: 2 to n + 1 line: n integers, these n integers form a number string.
Output format
an integer, which represents the length of the longest repeated substring.
Sample input
8 2
1 2 3 2 3 2 3 1
Sample output
4
Data size and convention
0 ≤ n ≤ 20000, 2 ≤ k ≤ n, 0 ≤ integer in number string ≤ 1000000
Suffix array:
- Use suff[i ]: to indicate the suffix starting with the i-th position.
- Suffix array sa[i] : Indicates the position of the suffix with rank i in the original string after all suffixes are sorted.
- First name group rank[i] : Indicates the current rank of the i-th place in the original string after all suffixes are sorted.
The relationship between the three:
Find the suffix array:
There are generally two ways to construct the sa array:
- Multiplication algorithm: O(nlogn)
- DC3 algorithm: O(n)
The algorithm will use the radix sorting
multiplication algorithm and the radix sorting can be found online.
LCP-longest common prefix:
height[i] : Represents the largest common prefix of suff[sa[i]] and suff[sa[i−1]], that is, the longest common prefix of two adjacent suffixes after ranking.
Illustration:
Application of suffix array:
1. The longest repeated substring can be overlapped
Given a string, find the longest repeated substring, these two substrings can overlap
2. The longest repeating substring cannot be overlapped
Given a string, find the longest repeating substring, these two substrings cannot overlap
3. The longest repeating substring of K times that can be overlapped
Given a string, find the longest repeating substring that appears at least K times, and these K substrings can overlap
Limited ability, you can understand by yourself : suffix array-a powerful tool for processing strings
Source program of this question:
#include<iostream>
using namespace std;
#define maxsize 100000
int Rank[maxsize], sa[maxsize];
int height[maxsize];
int sec[maxsize], t[maxsize];
int s[maxsize];//接收用户输入的数组
int num = maxsize;
int len, number;//长度和次数
inline void SA() //获得sa[]数组
{
for (int i = 1; i <= num; i++) t[i] = 0;
for (int i = 1; i <= len; i++) ++t[Rank[i] = s[i]];
for (int i = 1; i <= num; i++) t[i] += t[i - 1];
for (int i = len; i >= 1; i--) sa[t[Rank[i]]--] = i;
for (int k = 1; k <= len; k++)
{
int cnt = 0;
for (int i = len - k + 1; i <= len; i++) sec[++cnt] = i;
for (int i = 1; i <= len; i++) if (sa[i] > k) sec[++cnt] = sa[i] - k;
for (int i = 1; i <= num; i++) t[i] = 0;
for (int i = 1; i <= len; i++) ++t[Rank[i]];
for (int i = 1; i <= num; i++) t[i] += t[i - 1];
for (int i = len; i >= 1; i--) sa[t[Rank[sec[i]]]--] = sec[i], sec[i] = 0;
swap(Rank, sec);
Rank[sa[1]] = 1, cnt = 1;
for (int i = 2; i <= len; i++)
Rank[sa[i]] = (sec[sa[i]] == sec[sa[i - 1]] && sec[sa[i] + k] == sec[sa[i - 1] + k]) ? cnt : ++cnt;
if (cnt == len) break;
num = cnt;
}
}
void Getheight() //获得height[]数组
{
int j, k = 0;
for (int i = 1; i <= len; i++)
{
if (k) k--;
int j = sa[Rank[i] - 1];
while (s[i + k] == s[j + k]) k++;
height[Rank[i]] = k;
}
}
bool check(int mid)//利用height[]数组进行分类，然后统计组内是否满足重复的次数。
{
int group = 0;
for (int i = 1; i <= len; i++)
{
if (height[i] >= mid)
group++;
if (group >= number)
return true;
if (height[i] < mid)
group = 1;
}
return false;
}
int main()
{
cin >> len >> number;
for (int i = 1; i <= len; i++)
cin >> s[i];
SA();
Getheight();
int front=1, rear=len;
int maxlen = 0;
while (front <= rear)//二分法！
{
int mid = (front + rear) / 2;
if (check(mid))
{
front = mid + 1;
maxlen = mid;
}
else
rear = mid - 1;
}
cout << maxlen << endl;
}
Evaluation results:
