Maximum repetition substring(POJ - 3693)(sa(后缀数组)+st表)

The repetition number of a string is defined as the maximum number \(R\) such that the string can be partitioned into \(R\) same consecutive substrings. For example, the repetition number of "ababab" is 3 and "ababa" is 1.

Given a string containing lowercase letters, you are to find a substring of it with maximum repetition number.

Input

The input consists of multiple test cases. Each test case contains exactly one line, which
gives a non-empty string consisting of lowercase letters. The length of the string will not be greater than 100,000.

The last test case is followed by a line containing a '#'.

Output

For each test case, print a line containing the test case number( beginning with 1) followed by the substring of maximum repetition number. If there are multiple substrings of maximum repetition number, print the lexicographically smallest one.

Sample Input

ccabababc
daabbccaa
#

Sample Output

Case 1: ababab
Case 2: aa

题意：

找出原字符串中的重复次数最多的连续重复子串。

题解：

记这个连续重复子串为\(L\)，我们可以发现，这个字符串一定会覆盖\(s[0],s[L],s[L*2]\).....这些点中相邻的两个（因为长度至少为\(2L\)嘛）。假设它覆盖的是\(s[L*i]\)和\(s[L*(i+1)]\)，那么我们就往前和往后计算能匹配多远（往后匹配用到了后缀数组的height数组，往前匹配可以while到\(s[L*(i-1)]\)，越过\(s[L*(i-1)]\)的情况和前面计算的重复了，可以不算）

记往前匹配和往后匹配的最长长度为k，则重复次数为\(k/L+1\)。

再求lcp的时候用st表优化一下即可

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1000010;
char s[N];
int n;
int fir[N],sec[N],rnk[N],t[N],sa[N],b[N];
double log2(double x){
	return log(x)/log(2.0);
}
void msort(){
	memset(t,0,sizeof t);
	for(int i=1;i<=n;++i)t[sec[i]]++;
	for(int i=1;i<N;++i)t[i]+=t[i-1];
	for(int i=n;i;--i)b[t[sec[i]]--]=i;
	memset(t,0,sizeof t);
	for(int i=1;i<=n;++i)t[fir[b[i]]]++;
	for(int i=1;i<N;++i)t[i]+=t[i-1];
	for(int i=n;i;--i)sa[t[fir[b[i]]]--]=b[i];
}
int height[N];
void get_height(char *s){
	int k=0;
	for(int i=1;i<=n;++i){
		if(rnk[i]==1){
		    height[i]=0;
		    continue;
	    }
		if(k)--k;
	    int j=sa[rnk[i]-1];
	    while(i+k<=n&&j+k<=n&&s[i+k]==s[j+k])k++;
	    height[i]=k;
	}
}
int mn[N][20];
void get_height_st(){
	for(int i=1;i<=n;++i)mn[i][0]=height[sa[i]];
	int t=log2(n);
	for(int i=1;i<=t;++i){
		for(int j=1;j<=n;++j){
			if(j+(1<<(i-1))>n)mn[j][i]=mn[j][i-1];
			else mn[j][i]=min(mn[j][i-1],mn[j+(1<<(i-1))][i-1]);
		}
	}
}
int height_query(int l,int r){
	l=rnk[l],r=rnk[r];
	if(l>r)swap(l,r);l++;
	int t=log2(r-l+1);
	return min(mn[l][t],mn[r-(1<<t)+1][t]);
}
void get_sa(char *s){
	for(int i=1;i<=n;++i)rnk[i]=s[i];
	for(int k=1;k<=n;k*=2){
		for(int i=1;i<=n;++i){
			fir[i]=rnk[i];
			if(i+k>n)sec[i]=0;
			else sec[i]=rnk[i+k];
		}
		msort();
		int num=1;rnk[sa[1]]=1;
		for(int i=2;i<=n;++i){
			if(fir[sa[i]]!=fir[sa[i-1]]||sec[sa[i]]!=sec[sa[i-1]])num++;
			rnk[sa[i]]=num;
		}
		if(num==n)break;
	}	
}
int maxn,pos,len;
void find(){
	maxn=1;
	for(int i=1;i<=n/2;++i){
		for(int j=1;j+i<=n;j+=i){
			if(s[j]!=s[j+i])continue;
			int k=height_query(j,j+i),now,r;
			now=k/i+1,r=i-k%i;
			int cnt=0,p=j;
			for(int m=j-1;m>j-i&&s[m]==s[m+i]&&m;--m){
				cnt++;
				if(cnt==r)now++,p=m;
				else p=rnk[p]>rnk[m]?m:p;
			}
			if(now>maxn)maxn=now,pos=p,len=i;
			else if(now==maxn&&rnk[pos]>rnk[p])pos=p,len=i;
		}
	}
}
int main(){
    int js=0;
	while(1){
		js++;
		cin>>s+1;
		if(s[1]=='#')break;
		n=strlen(s+1);
		get_sa(s);
		get_height(s);
		get_height_st();
		find();
		printf("Case %d: ",js);
		for(int i=pos;i<=pos+len*maxn-1;++i){
			putchar(s[i]);
		}puts("");
	}
}