poj 3693 后缀数组 重复次数最多的连续重复子串

Maximum repetition substring

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10612		Accepted: 3277

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：

给定一个字符串，求重复次数最多的连续重复子串。输出字典序最小的。

代码：

//论文题，还是挺难的。连续出现1次是肯定的，我们枚举重复部分的长度L，把然后把字符串分成长为L的若干段，s[0],s[L],s[2*L]...
//那么要求的子串必然会包含相邻的两段，所以看s[i*L]和s[i*L+L]向前和向后能匹配多远（i*L~i*L+L是其中的一部分），即求出来他们
//的lcp，向后最远最多是lcp的终点，向前我们可以给他向前枚举L长度个单位（没必要枚举超过L，前一段计算包含了）看看最多能匹
//配多少，还要保持字典序最小。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=200000;
int sa[MAXN+9],he[MAXN+9],ra[MAXN+9],xx[MAXN+9],yy[MAXN+9],buc[MAXN+9],f[MAXN+9][30];
char s[MAXN+9];
int len,m;
void get_suf()
{
    int *x=xx,*y=yy;
    for(int i=0;i<m;i++) buc[i]=0;
    for(int i=0;i<len;i++) buc[x[i]=s[i]]++;
    for(int i=1;i<m;i++) buc[i]+=buc[i-1];
    for(int i=len-1;i>=0;i--) sa[--buc[x[i]]]=i;
    for(int k=1;k<=len;k<<=1){
        int p=0;
        for(int i=len-1;i>=len-k;i--) y[p++]=i;
        for(int i=0;i<len;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
        for(int i=0;i<m;i++) buc[i]=0;
        for(int i=0;i<len;i++) buc[x[y[i]]]++;
        for(int i=1;i<m;i++) buc[i]+=buc[i-1];
        for(int i=len-1;i>=0;i--) sa[--buc[x[y[i]]]]=y[i];
        swap(x,y);
        p=1;x[sa[0]]=0;
        for(int i=1;i<len;i++){
            if(y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k])
                x[sa[i]]=p-1;
            else x[sa[i]]=p++;
        }
        if(p>=len) break;
        m=p;
    }
    for(int i=0;i<len;i++) ra[sa[i]]=i;
    int k=0;
    for(int i=0;i<len;i++){
        if(ra[i]==0) { he[0]=0; continue; }
        if(k) k--;
        int j=sa[ra[i]-1];
        while(s[i+k]==s[j+k]&&i+k<len&&j+k<len) k++;
        he[ra[i]]=k;
    }
}
void rmq1()
{
    he[0]=he[len]=0;
    for(int i=0;i<=len;i++) f[i][0]=he[i];
    for(int j=1;(1<<j)<=len;j++){
        for(int i=0;i+(1<<j)-1<=len;i++)
            f[i][j]=min(f[i][j-1],f[i+(1<<(j-1))][j-1]);
    }
}
int rmq2(int ql,int qr)
{
    if(ql>qr) swap(ql,qr);
    ql++;
    int k=0;
    while(1<<(k+1)<=qr-ql+1) k++;
    return min(f[ql][k],f[qr-(1<<k)+1][k]);
}
void solve(int &st,int &ed,int ans)
{
    for(int l=1;l*2<=len;l++){
        for(int i=0;i+l<len;i+=l){
            if(s[i]!=s[i+l]) continue;
            int ll=rmq2(ra[i],ra[i+l]);
            for(int j=0;j<=l;j++){
                int ii=i-j;
                if(ii<0||s[ii]!=s[ii+l]) break;
                int num=(j+ll)/l+1;
                if(num>ans||(num==ans&&ra[st]>ra[ii])){
                    ans=num;
                    st=ii;ed=st+ans*l-1;
                }
            }
        }
    }
}
int main()
{
    int cas=0;
    while(scanf("%s",s)&&s[0]!='#'){
        len=strlen(s);
        m=200;
        get_suf();
        rmq1();
        int st=sa[0],ed=sa[0];
        solve(st,ed,1);
        printf("Case %d: ",++cas);
        for(int i=st;i<=ed;i++) printf("%c",s[i]);
        printf("
");
    }
    return 0;
}