poj3693Maximum repetition substring【后缀数组+RMQ求重复最多连续子串】

Description

The repetition number of a string is defined as the maximum number R such that the string can be partitioned intoR 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

ccabababcdaabbccaa#

Sample Output

Case 1: abababCase 2: aa

Source

2008 Asia Hefei Regional Contest Online by USTC

花费了一晚上本来要学linux和php的时间又钻研后缀数组，可算整明白点了，值~

对于数组的解释：x，y数组都是记录的以当前长度排列的暂时rank值，height[]数组的下标是排名，对于大多数后缀数组的题为何用rmq？为了求以j,k开头的最长公共子序列，min(height[rank[j]+1],height[rank[j]+2]……height[rank[k]])，因为是“公共”所以取min,因为是区间最小，所以用RMQ

这个题要求重复最多的连续子串，那么我们先假设一个重复子串单元为L,从数组开头开始遍历，每次拿出i,i+L求LCP，那么LCP包含了至少LCP/L个循环单元，还得记得i,i+L是按着的，重复次数还得加1.这还没完，还要处理余数的问题。处理之后所有可能重复最小单元长度值都放到a[]数组里，我们遍历sa数组，这个数组本来就是以字典序存储的，那么遇到的第一个符合条件的就输出

/*******************poj36932016.2.2110112K313MSG++3164B*******************/#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;const int MAXN=100010;int sa[MAXN];int t1[MAXN],t2[MAXN],c[MAXN];int rank[MAXN],height[MAXN];void build_sa(int s[],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]=s[i]]++;    for(i=1;i<m;i++)c[i]+=c[i-1];    for(i=n-1;i>=0;i--)sa[--c[x[i]]]=i;    for(j=1;j<=n;j<<=1)    {        p=0;        for(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++)c[i]=0;        for(i=0;i<n;i++)c[x[y[i]]]++;        for(i=1;i<m;i++)c[i]+=c[i-1];        for(i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i];        swap(x,y);        p=1;x[sa[0]]=0;        for(i=1;i<n;i++)            x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;        if(p>=n)break;        m=p;    }}void getHeight(int s[],int n){    int i,j,k=0;    for(i=0;i<=n;i++)rank[sa[i]]=i;    for(i=0;i<n;i++)    {        if(k)k--;        j=sa[rank[i]-1];        while(s[i+k]==s[j+k])k++;        height[rank[i]]=k;    }}int mm[MAXN];int best[20][MAXN];void initRMQ(int n){    mm[0]=-1;    for(int i=1;i<=n;i++)        mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];    for(int i=1;i<=n;i++)best[0][i]=i;    for(int i=1;i<=mm[n];i++)        for(int j=1;j+(1<<i)-1<=n;j++)        {            int a=best[i-1][j];            int b=best[i-1][j+(1<<(i-1))];            if(height[a]<height[b])best[i][j]=a;            else best[i][j]=b;        }}int askRMQ(int a,int b){    int t;    t=mm[b-a+1];    b-=(1<<t)-1;    a=best[t][a];b=best[t][b];    return height[a]<height[b]?a:b;}int lcp(int a,int b){    a=rank[a];b=rank[b];    if(a>b)swap(a,b);    return height[askRMQ(a+1,b)];}char str[MAXN];int a[MAXN],r[MAXN],iCase;int main(){    //freopen("cin.txt","r",stdin);    iCase=1;    while(~scanf("%s",str))    {        if(str[0]=='#') break;        int n=strlen(str);        for(int i=0;i<=n;i++) r[i]=str[i];        build_sa(r,1+n,128);        getHeight(r,n);        initRMQ(n);        int cnt=0,maxn=0,step;        for(int l=1;l<n;l++)        {            for(int i=0;i+l<n;i+=l)            {                int t1=lcp(i,i+l);                int re=t1%l;                step=t1/l+1;                int j=i-(l-t1%l);                if(j>=0&&re&&lcp(j,j+l)>=t1) step++;                if(step>maxn)                {                    maxn=step;                    cnt=0;                    a[cnt++]=l;                }                else if(step==maxn)                {                    a[cnt++]=l;                }            }        }        int len=-1,st;        for(int i=1;i<=n&&len==-1;i++)//i表示排名        {            for(int j=0;j<cnt;j++)            {                if(lcp(sa[i],sa[i]+a[j])>=(maxn-1)*a[j])                {                    len=a[j];                    st=sa[i];                    break;                }            }        }        str[maxn*len+st]=0;        printf("Case %d: %s\n",iCase++,str+st);    }    return 0;}