Maximum repetition substring (poj3693 后缀数组求重复次数最多的连续重复子串）

最新推荐文章于 2020-10-20 21:19:05 发布

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最新推荐文章于 2020-10-20 21:19:05 发布

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原文链接：http://www.cnblogs.com/ERKE/p/3598368.html

本文介绍了一种寻找字符串中具有最大重复次数子串的高效算法。该算法首先定义了重复数的概念，即一个字符串可以被划分为相同连续子串的最大数量。通过使用字符串处理技术和数据结构，如后缀数组和高度数组，来确定输入字符串中的最大重复子串。此外，还提供了一个C++实现示例。

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Maximum repetition substring

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6328		Accepted: 1912

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

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <math.h>
  4 #include <vector>
  5 #include <string.h>
  6 using namespace std;
  7 #define N 100005
  8 char a[N];
  9 int c[N],d[N],e[N],sa[N],height[N],n,b[N],m,dp[N][21];
 10 int cmp(int *r,int a,int b,int l)
 11 {
 12     return r[a]==r[b]&&r[a+l]==r[b+l];
 13 }
 14 void da()
 15 {
 16     int i,j,p,*x=c,*y=d,*t;
 17     memset(b,0,sizeof(b));
 18     for(i=0; i<n; i++)b[x[i]=a[i]]++;
 19     for(i=1; i<m; i++)b[i]+=b[i-1];
 20     for(i=n-1; i>=0; i--)sa[--b[x[i]]]=i;
 21     for(j=1,p=1; p<n; j*=2,m=p)
 22     {
 23         for(p=0,i=n-j; i<n; i++)y[p++]=i;
 24         for(i=0; i<n; i++)if(sa[i]>=j)y[p++]=sa[i]-j;
 25         for(i=0; i<n; i++)e[i]=x[y[i]];
 26         for(i=0; i<m; i++)b[i]=0;
 27         for(i=0; i<n; i++)b[e[i]]++;
 28         for(i=1; i<m; i++)b[i]+=b[i-1];
 29         for(i=n-1; i>=0; i--)sa[--b[e[i]]]=y[i];
 30         for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)
 31             x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
 32     }
 33 }
 34 void callheight()
 35 {
 36     int i,j,k=0;
 37     b[0]=0;
 38     for(i=1; i<n; i++)b[sa[i]]=i;
 39     for(i=0; i<n-1; height[b[i++]]=k)
 40         for(k?k--:0,j=sa[b[i]-1]; a[i+k]==a[j+k]; k++);
 41 }
 42 int fun(int i,int j)
 43 {
 44     i=b[i];
 45     j=b[j];
 46     if(i>j)swap(i,j);
 47     i++;
 48     int k=(int)(log(j-i+1.0)/log (2.0));
 49     return min(dp[i][k],dp[j-(1<<k)+1][k]);
 50 }
 51 void initrmq()
 52 {
 53     int i,j;
 54     memset(dp,0,sizeof(dp));
 55     for(i=0; i<=n; i++)
 56         dp[i][0]=height[i];
 57     for(j=1; (1<<j)<=n; j++)
 58         for(i=0; i+(1<<j)<=n; i++)
 59             dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
 60 }
 61 int main()
 62 {
 63     int t=1,i,j,yy;
 64     while(~scanf("%s",a))
 65     {
 66         yy=0;
 67         if(a[0]=='#')break;
 68         n=strlen(a);
 69         n++;
 70         m=130;
 71         da();
 72         callheight();
 73         initrmq();
 74         memset(c,0,sizeof(c));
 75         int max=1;
 76         n--;
 77         for(i=1; i<n/2; i++)
 78         {
 79             for(j=0; j+i<n; j+=i)
 80             {
 81                 int k=fun(j,j+i);
 82                 int kk=k/i+1;
 83                 int tt=i-k%i;
 84                 tt=j-tt;
 85                 if (tt>=0&&k%i!=0)
 86                     if(fun(tt,tt+i)>=k)
 87                         kk++;
 88                 if(max<kk)
 89                 {
 90                     yy=0;
 91                     c[yy++]=i;
 92                     max=kk;
 93                 }
 94                 else if(max==kk)
 95                 {
 96                     c[yy++]=i;
 97                 }
 98             }
 99         }
100         printf("Case %d: ",t++);
101         int sta,m;
102         for(i=1; i<n; i++)
103         {
104             for(j=0; j<yy; j++)
105             {
106 
107                 if(fun(sa[i],sa[i]+c[j])>=(max-1)*c[j])
108                 {
109                     sta=sa[i];
110                     m=max*c[j];
111                     break;
112                 }
113             }
114             if(j<yy)break;
115         }
116         for(i=0; i<m; i++)putchar(a[sta+i]);
117         printf("\n");
118     }
119 }