题意：
给出一串数，求可重叠k次的最长重复子串。
思路：
后缀数组。
先二分答案，然后将后缀分成若干组。
这里要判断的是有没有一个组的后缀个数不小于k。
如果有，那么存在k个相同的子串满足条件，否则不存在。
时间复杂度为O(nlogn) 。
可参考 09 年国家队论文 罗穗骞《后缀数组——处理字符串的有力工具》
代码：
#include <stdio.h> #include <algorithm> #include <iostream> #include <string.h> #include <queue> using namespace std; /*suffix array *倍增算法 O(n*logn) *待排序数组长度为n,放在0~n-1中，在最后面补一个0 *build_sa(,n+1, ); //注意是n+1; *getHeight(,n); *例如： *n = 8; *num[] = { 1, 1, 2, 1, 1, 1, 1, 2, $ }; 注意num最后一位为0，其他大于0 *rank[] = { 4, 6, 8, 1, 2, 3, 5, 7, 0 }; rank[0~n-1]为有效值，rank[n]必定为0无效值 *sa[] = { 8, 3, 4, 5, 0, 6, 1, 7, 2 }; sa[1~n]为有效值，sa[0]必定为n是无效值 *height[]= { 0, 0, 3, 2, 3, 1, 2, 0, 1 }; height[2~n]为有效值 * */ const int MAXN=50010; int sa[MAXN],m; //SA数组，表示将S的n个后缀从小到大排序后把排好序的 //的后缀的开头位置顺次放入SA中 int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值 int rank[MAXN],height[MAXN]; //待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m, //除s[n-1]外的所有s[i]都大于0，r[n-1]=0 //函数结束以后结果放在sa数组中 void build_sa(int s[],int n,int m) { int i,j,p,*x=t1,*y=t2; //第一轮基数排序，如果s的最大值很大，可改为快速排序 for(i=0; i<m; i++)c[i]=0; for(i=0; i<n; i++)c[x[i]=s[i]]++; for(i=1; i<m; i++)c[i]+=c[i-1]; for(i=n-1; i>=0; i--)sa[--c[x[i]]]=i; for(j=1; j<=n; j<<=1) { p=0; //直接利用sa数组排序第二关键字 for(i=n-j; i<n; i++)y[p++]=i; //后面的j个数第二关键字为空的最小 for(i=0; i<n; i++)if(sa[i]>=j)y[p++]=sa[i]-j; //这样数组y保存的就是按照第二关键字排序的结果 //基数排序第一关键字 for(i=0; i<m; i++)c[i]=0; for(i=0; i<n; i++)c[x[y[i]]]++; for(i=1; i<m; i++)c[i]+=c[i-1]; for(i=n-1; i>=0; i--)sa[--c[x[y[i]]]]=y[i]; //根据sa和x数组计算新的x数组 swap(x,y); p=1; x[sa[0]]=0; for(i=1; i<n; i++) x[sa[i]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++; if(p>=n)break; m=p;//下次基数排序的最大值 } int k = 0; n--; for(i = 0; i <= n; i++) rank[sa[i]] = i; for(i = 0; i < n; i++) { if(k) k--; j = sa[rank[i]-1]; while(s[i+k] == s[j+k]) k++; height[rank[i]] = k; } } bool judge(int mid,int n){ for(int i=2;i<=n;i++){ int cut=0; while(height[i]>=mid&&i<=n){ cut++; i++; } if(cut>=m-1) return 1; } return 0; } int solve(int n){ int le=1,ri=n,mid,ans=-1; while(ri-le>1){ mid=(le+ri)/2; if(judge(mid,n)){ le=mid; ans=mid; }else ri=mid; } return ans; } int main(){ int n; int s[MAXN]; while(scanf("%d%d",&n,&m)!=-1){ for(int i=0;i<n;i++) scanf("%d",&s[i]); build_sa(s,n+1,20010); printf("%d\n",solve(n)); } }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 Ktimes.
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