poj 3693 Maximum repetition substring 重复次数最多的子串(若有多个 输出字典序最小的子串) 后缀数组 (DC3) (SA）

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

 

 

 

//

 

 

 

//

#include <iostream>
using namespace std;

#define maxn 6100000
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
int wa[maxn],wb[maxn],wv[maxn],wss[maxn];
int c0(int *r,int a,int b)
{return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2];}
int c12(int k,int *r,int a,int b)
{if(k==2) return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1);
else return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1];}
void sort(int *r,int *a,int *b,int n,int m)
{
     int i;
     for(i=0;i<n;i++) wv[i]=r[a[i]];
     for(i=0;i<m;i++) wss[i]=0;
     for(i=0;i<n;i++) wss[wv[i]]++;
     for(i=1;i<m;i++) wss[i]+=wss[i-1];
     for(i=n-1;i>=0;i--) b[--wss[wv[i]]]=a[i];
     return;
}
void dc3(int *r,int *sa,int n,int m)
{
     int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p;
     r[n]=r[n+1]=0;
     for(i=0;i<n;i++) if(i%3!=0) wa[tbc++]=i;
     sort(r+2,wa,wb,tbc,m);
     sort(r+1,wb,wa,tbc,m);
     sort(r,wa,wb,tbc,m);
     for(p=1,rn[F(wb[0])]=0,i=1;i<tbc;i++)
     rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++;
     if(p<tbc) dc3(rn,san,tbc,p);
     else for(i=0;i<tbc;i++) san[rn[i]]=i;
     for(i=0;i<tbc;i++) if(san[i]<tb) wb[ta++]=san[i]*3;
     if(n%3==1) wb[ta++]=n-1;
     sort(r,wb,wa,ta,m);
     for(i=0;i<tbc;i++) wv[wb[i]=G(san[i])]=i;
     for(i=0,j=0,p=0;i<ta && j<tbc;p++)
     sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++];
     for(;i<ta;p++) sa[p]=wa[i++];
     for(;j<tbc;p++) sa[p]=wb[j++];
     return;
}
int rank[maxn],height[maxn];
void calheight(int *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 mm[maxn];
int best[20][maxn];
void initRMQ(int n,int* RMQ)//这个RMQ是从1..n的 但是height碰巧也是从1开始
{
     int i,j,a,b;
     for(mm[0]=-1,i=1;i<=n;i++)
     mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
     for(i=1;i<=n;i++) best[0][i]=i;
     for(i=1;i<=mm[n];i++)
     for(j=1;j<=n+1-(1<<i);j++)
     {
       a=best[i-1][j];
       b=best[i-1][j+(1<<(i-1))];
       if(RMQ[a]<RMQ[b]) best[i][j]=a;//最小RMQ
       else best[i][j]=b;
     }
     return;
}
int askRMQ(int a,int b,int* RMQ)
{
    int t;
    t=mm[b-a+1];b-=(1<<t)-1;
    a=best[t][a];b=best[t][b];
    return RMQ[a]<RMQ[b]?a:b;//最小RMQ
}
int lcp(int a,int b)
{
    int t;
    a=rank[a];b=rank[b];
    if(a>b) {t=a;a=b;b=t;}
    return height[askRMQ(a+1,b,height)];
}
/////////用法
//    dc3(r,sa,n+1,128);//转int后的字符串 r[0]=0 sa数组存放 长度+1 字符集大小
//    calheight(r,sa,n);//r同上 sa为上面返回的 n就是长度
//    所有数组都要开到3倍以上 如果是拼接串 要开到总长3倍
int r[maxn],sa[maxn];
int n;
char s[maxn]="bacbacbaba";
int maxrep,beg,len;
void work(int k)
{
int i,j,l,r;
for (i=0;i+k<n;i+=k) if (s[i]==s[i+k])
{
   j=i+k;
   r=lcp(i,j);
   l=lcp(n*2-i,n*2-j);
   int lt,rt;
   int rep;
   int thl;
   lt=i-l+1;rt=i+r+k;
   rep=(l+r-1)/k+1;
   thl=rep*k;

   if (l+r-1<k) continue;
   if (rep<maxrep) continue;

   for (j=lt;j+thl<=rt;j++)
    if (rep>maxrep||(rep==maxrep&&rank[j]<rank[beg])||(rep==maxrep&&rank[j]==rank[beg]&&thl<len))
    {
     maxrep=rep;
     beg=j;
     len=thl;
    }
}
}

int main()
{
int i,j;
int cas=0;
while (1)
{
   gets(s);
   if (s[0]=='#') break;
   cas++;
   printf("Case %d: ",cas);
   n=strlen(s);
   for (i=0;i<n;i++) r[i]=s[i]+1;
   r[n]=1;
   for (i=0;i<n;i++) r[i+n+1]=s[n-i-1]+1;
   r[n*2+1]=0;
   dc3(r,sa,n*2+2,300);
     calheight(r,sa,n*2+1);
   initRMQ(n*2+1,height);
   j=0;
   for (i=1;i<n;i++) if (s[i]<s[j]) j=i;
   maxrep=1;beg=j;len=1;
   for (i=1;i<n;i++) work(i);
   for (i=beg;i<beg+len;i++) printf("%c",s[i]);
   printf("/n");
}
return 0;
}

倍增算法：

http://blog.youkuaiyun.com/hqd_acm/article/details/6327103