POJ3693 Maximum repetition substring （后缀数组）

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：给一个字符串求最长重复子串。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 100005
#define maxn 100005
using namespace std;
//以下为倍增算法求后缀数组  
int wa[maxn],wb[maxn],wv[maxn],Ws[maxn];  
int cmp(int *r,int a,int b,int l)  
{return r[a]==r[b]&&r[a+l]==r[b+l];}  
void da(const char *r,int *sa,int n,int m){  
	int i,j,p,*x=wa,*y=wb,*t;   
	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(t=x,x=y,y=t,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++;   
	}   
	return;   
}  
int sa[maxn],Rank[maxn],height[maxn];  
//求height数组  
void calheight(const char *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++);  
	return;  
}
int dp[maxn][20];
void Rmq_Init(int n){
	int m=floor(log(n+0.0)/log(2.0));
	for(int i=1;i<=n;i++) dp[i][0]=height[i];
	for(int i=1;i<=m;i++){
		for(int j=n;j;j--){
			dp[j][i]=dp[j][i-1];
			if(j+(1<<(i-1))<=n)
				dp[j][i]=min(dp[j][i],dp[j+(1<<(i-1))][i-1]);
		}
	}
}
int Rmq_Query(int l,int r){
	int a=Rank[l],b=Rank[r];
	if(a>b) swap(a,b);
	a++;
	int m=floor(log(b-a+1.0)/log(2.0));
	return min(dp[a][m],dp[b-(1<<m)+1][m]);
}
char str[maxn];
int main(){
	int cas=0;
	while(scanf("%s",str)!=EOF&&str[0]!='#'){
		int n=strlen(str);
		da(str,sa,n+1,130);
		calheight(str,sa,n);
		Rmq_Init(n);
		int cnt=0,mmax=0,a[maxn];
		for(int l=1;l<n;l++){
			for(int i=0;i+l<n;i+=l){
				int r=Rmq_Query(i,i+l);
				int step=r/l+1;
				int k=i-(l-r%l);
				if(k>=0&&r%l)
			    	if(Rmq_Query(k,k+l)>=r) 
						step++;
				if(step>mmax){
					mmax=step;
					cnt=0;
					a[cnt++]=l;
				}
				else if(step==mmax)
					a[cnt++]=l;
			}
		}
		int len=-1,st;
		for(int i=1;i<=n&&len==-1;i++){
			for(int j=0;j<cnt;j++){
				int l=a[j];
				if(Rmq_Query(sa[i],sa[i]+l)>=(mmax-1)*l){
					len=l;
					st=sa[i];
					break;
				}
			}
		}
		printf("Case %d: ",++cas);
		for(int i=st,j=0;j<len*mmax;j++,i++) printf("%c",str[i]);
		printf("\n");
	}
	return 0;
}