Maximum repetition substring 后缀数组

Maximum repetition substring

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7578		Accepted: 2281

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

 

题意：给出一串字符，需要求这串字符中的最长重复子串，要是有多个，输出字典序最小的.........

 

 

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cstdlib>
using namespace std;
const int INF=0x5fffffff;
const double EXP=1e-8;
const int MS=200005;
//   KMP  TRIE  DFA  SUFFIX
int dp[MS][30];        //    RMQ
int t1[MS],t2[MS],c[MS],v[MS];
int rank[MS],sa[MS],height[MS];
char str[MS],str1[MS];
int s[MS];
int cmp(int *r,int a,int b,int k)
{
    return r[a]==r[b]&&r[a+k]==r[b+k];
}

void get_sa(int *r,int *sa,int n,int m)
{
    int i,j,p,*x=t1,*y=t2;
    for(i=0;i<m;i++)
        c[i]=0;
    for(i=0;i<n;i++)
        c[x[i]=r[i]]++;
    for(i=1;i<m;i++)
        c[i]+=c[i-1];
    for(i=n-1;i>=0;i--)
        sa[--c[x[i]]]=i;
    p=1;j=1;
    for(;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++)
            v[i]=x[y[i]];
        for(i=0;i<m;i++)
            c[i]=0;
        for(i=0;i<n;i++)
            c[v[i]]++;
        for(i=1;i<m;i++)
            c[i]+=c[i-1];
        for(i=n-1;i>=0;i--)
            sa[--c[v[i]]]=y[i];
        swap(x,y);
        x[sa[0]]=0;
        for(p=1,i=1;i<n;i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    }
}

void get_height(int *r,int n)
{
    int i,j,k=0;
    for(i=1;i<=n;i++)
        rank[sa[i]]=i;
    //height[i]>=height[i-1]-1;
    for(i=0;i<n;i++)
    {
        if(k)
            k--;
        else
            k=0;
        j=sa[rank[i]-1];
        while(r[i+k]==r[j+k])
            k++;
        height[rank[i]]=k;
    }
}

void rmq_init(int n)
{
    for(int i=1;i<=n;i++)  dp[i][0]=height[i];
    for(int j=1;(1<<j)<=n;j++)
    for(int i=1;i+(1<<j)-1<=n;i++)
    dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}

int rmq(int ll,int rr)
{
    int k=0;
    ll=rank[ll];
    rr=rank[rr];
    if(ll>rr)
    {
        int tmp=ll;
        ll=rr;
        rr=tmp;
    }
    ll++;
    while((1<<(k+1))<=rr-ll+1) k++;
    return min(dp[ll][k],dp[rr-(1<<k)+1][k]);
}

int main()
{
    int text=0;
    while(scanf("%s",str)>0)
    {
        if(str[0]=='#')
        break;
        int len=strlen(str);
        for(int i=0;i<len;i++)
        s[i]=str[i]-'a'+1;
        s[len]=0;
        get_sa(s,sa,len+1,30);
        get_height(s,len);
        rmq_init(len);
        int ans=0,pos=0,lenn;
        for(int i=1;i<=len/2;i++)
        {
            for(int j=0;j<len-i;j+=i)
            {
                if(str[j]!=str[j+i])
                continue;
                int k=rmq(j,j+i);
                int tol=k/i+1;
                //printf("%d
",tol);
                int r=i-k%i;
                int p=j;
                int cnt=0;
                for(int m=j-1;m>j-i&&str[m]==str[m+i]&&m>=0;m--)
                {
                    cnt++;
                    if(cnt==r)
                    {
                        tol++;
                        p=m;
                    }
                    else if(rank[p]>rank[m])
                    {
                        p=m;
                    }
                }
                if(ans<tol)
                {
                    ans=tol;
                    pos=p;
                    lenn=tol*i;
                }
                else if(ans==tol&&rank[pos]>rank[p])
                {
                    pos=p;
                    lenn=tol*i;
                }
            }
        }
        printf("Case %d: ",++text);
    //  printf("%d %d %d
",ans,pos,lenn);
        if(ans<2)                         //这里，如果字符总长度小于2，那么就在原串中找出一个最小的字符就好
        {
            char ch='z';
            for(int i=0;i<len;i++)
            if(str[i]<ch)
            ch=str[i];
            printf("%c
",ch);
            continue;
        }
        for(int i=pos;i<pos+lenn;i++)
        printf("%c",str[i]);
        printf("
");
    }
    return 0;
}