poj3261Milk Patterns(可重叠的k次最长重复子串)
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 12401 | Accepted: 5490 | |
| Case Time Limit: 2000MS | ||
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
Lines 2.. N+1: N integers, one per line, the quality of the milk on day i appears on the ith line.
Output
Sample Input
8 2 1 2 3 2 3 2 3 1
Sample Output
4
Source
先二分答案，然后将后缀分成若干组。不
同的是，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。这个做法的时间复杂度为 O(nlogn)。
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
const int maxn=20000+100;
int n,k;
struct node{
int num;
int id;
}a[maxn];
int s[maxn];
int sa[maxn],t[maxn],t2[maxn],c[maxn];
//构造字符串s的后缀数组,每个字符值必须为0~m-1
void build_sa(int m){
int *x=t,*y=t2;
//基数排序
for(int i=0;i<m;i++) c[i]=0;
for(int i=0;i<n;i++) c[x[i]=s[i]]++;
for(int i=1;i<m;i++) c[i]+=c[i-1];
for(int i=n-1;i>=0;i--) sa[--c[x[i]]]=i;
for(int k=1;k<=n;k<<=1){
int p=0;
//直接利用sa数组排序第二关键字
for(int i=n-k;i<n;i++) y[p++]=i;
for(int i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
//基数排序第一关键字
for(int i=0;i<m;i++) c[i]=0;
for(int i=0;i<n;i++) c[x[y[i]]]++;
for(int i=1;i<m;i++) c[i]+=c[i-1];
for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
//根据sa和y计算新的x数组
swap(x,y);
p=1;
x[sa[0]]=0;
for(int i=1;i<n;i++)
x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;
if(p>=n)
break;
m=p; //下次基数排序的最大值
}
}
int rank1[maxn],height[maxn];
void getHeight(){
int i,j,k=0;
for(i=0;i<=n;i++) rank1[sa[i]]=i;
for(i=0;i<n;i++){
if(k)
k--;
int j=sa[rank1[i]-1];
while(s[i+k]==s[j+k]) k++;
height[rank1[i]]=k;
}
}
bool check(int m){
int cnt=0;
for(int i=0;i<=n;i++){
if(height[i]>=m){
cnt++;
if(cnt>=k-1)
return true;
}
else
cnt=0;
}
return false;
}
bool cmp(node b,node d){
return b.num<d.num;
}
int main(){
while(scanf("%d%d",&n,&k)!=EOF){
for(int i=0;i<n;i++){
scanf("%d",&a[i].num);
a[i].id=i;
a[i].num++;
}
sort(a,a+n,cmp);
int cnt=1;
s[a[0].id]=1;
for(int i=1;i<n;i++){
if(a[i].num==a[i-1].num)
s[a[i].id]=s[a[i-1].id];
else
s[a[i].id]=s[a[i-1].id]+1;
}
//for(int i=0;i<n;i++)
//printf("%d\n",s[i]);
s[n++]=0;
build_sa(n);
n--;
getHeight();
int low=1,high=n;
while(high-low>=0){
int mid=(low+high)>>1;
if(check(mid))
low=mid+1;
else
high=mid-1;
}
printf("%d\n",high);
}
return 0;
}
/*
4 2
1 3 1 4
8 2
1 2 3 2 3 2 3 1
*/