poj 3693 Maximum repetition substring //后缀数组

Maximum repetition substring

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2233		Accepted: 547

Description

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

Source

2008 Asia Hefei Regional Contest Online by USTC

 

/* 我的这种算法实际上是0(n^2)的，但是实际远远小于这个 首先呢，要保存所有可能取到最大重复值的长度 接着就要用到sa数组了 sa数组是保证字典序最好的利器 第一个循环是sa数组从1到n 接着第二个循环是可行长度的枚举 只要出现 lcp(ts,ts+tl)>=(max-1)*tl ts=sa[i] tl=枚举的长度，max代表最大重复 就找到了 */ #include<math.h> #include<stdio.h> #include<cstring> using namespace std; const int maxn=100010; const int inf=5000000; int w[maxn],wa[maxn],wb[maxn],wv[maxn]; int sa[maxn],rank[maxn],height[maxn]; int a[maxn],f[maxn][20],n,ft[maxn],len[maxn]; int num; char s[maxn]; int cmp(int *r,int a,int b,int l) { return r[a]==r[b]&&r[a+l]==r[b+l]; } void da(int *r,int *sa,int n,int m) { int i,j,p,*x=wa,*y=wb,*t; for (i=0; i<m; i++) w[i]=0; for (i=0; i<n; i++) w[x[i]=r[i]]++; for (i=1; i<m; i++) w[i]+=w[i-1]; for (i=n-1; i>=0; i--) sa[--w[x[i]]]=i; for (p=1,j=1; p<n; m=p,j*=2) { 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<m; i++) w[i]=0; for (i=0; i<n; i++) w[wv[i]=x[y[i]]]++; for (i=1; i<m; i++) w[i]+=w[i-1]; for (i=n-1; i>=0; i--) sa[--w[wv[i]]]=y[i]; for (t=x,x=y,y=t,x[sa[0]]=0,p=1,i=1; i<n; i++) x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++; } return; } void cal(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 nmin(int a,int b) { return a<b?a:b; } void rmq(int n) { int i,j; for (i=1; i<=n; i++) f[i][0]=height[i]; for (j=1; j<20; j++) for (i=1; i+(1<<j)-1<=n; i++) f[i][j]=nmin(f[i][j-1],f[i+(1<<j-1)][j-1]); return; } int lcp(int a,int b) { int x=rank[a],y=rank[b]; if (x>y) { int t=x; x=y; y=t; } x++; int t=ft[y-x+1]; return nmin(f[x][t],f[y-(1<<t)+1][t]); } int main() { int cas=1; int i; char x; for (i=0; i<maxn; i++) ft[i]=int(double(log(i*1.00))/log(2.00)); while (scanf("%s",s)!=EOF) { if(s[0]=='#') break; n=strlen(s); for (i=0; i<n; i++) a[i]=(int)s[i]; a[n]=0; da(a,sa,n+1,128); cal(a,sa,n); rmq(n); int k,max=0,r=0,t; num=0; for(int l=1; l<n; l++)//枚举长度 for(int i=0; i+l<n; i+=l) { k=lcp(i,i+l); r=k/l+1;//注意为什么是k/l+1 t=i-(l-k%l); if (t>=0&&k%l!=0) if (lcp(t,t+l)>=k) r++; if (r>max) { num=0; len[++num]=l; max=r; } else if(r==max) { len[++num]=l; } //printf("l=%d r=%d i=%d i+l=%d/n",l,r,i,i+l); } // printf("%d/n",max); int start,l; for(int i=1;i<n;++i)//sa应从1开始，因为sa[0]是0 { int ts=sa[i];//枚举这个开始 for(int j=1;j<=num;++j) { int tl=len[j]; if(lcp(ts,ts+tl)>=(max-1)*tl) { start=ts; l=tl*max; i=n; break; } } } printf("Case %d: ", cas++); for (int i = 0; i < l; i++) printf("%c", s[start+ i]); printf("/n"); } return 0; }