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

    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("
    ");
    118     }
    119 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ERKE/p/3598368.html
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