POJ 3693 Maximum repetition substring

lcp也上场了。。。
断断续续写了几天。。。。最后被后缀数组一套带走了。。。。。
就当是后缀数组的模板吧。。。  

Maximum repetition substring

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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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

ccabababcdaabbccaa#

Sample Output

Case 1: abababCase 2: aa

Source

2008 Asia Hefei Regional Contest Online by USTC

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#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <algorithm>using namespace std;const int maxn=1000012;int sa[maxn],rank[maxn],rank2[maxn],h[maxn],c[maxn],*x,*y,ans[maxn];char str[maxn];bool cmp(int* r,int a,int b,int l,int n){    if(r[a]==r[b]&&a+l<n&&b+l<n&&r[a+l]==r[b+l]) return true;    return false;}void radix_sort(int n,int sz){    for(int i=0;i<sz;i++) c[i]=0;    for(int i=0;i<n;i++) c[x[y[i]]]++;    for(int i=1;i<sz;i++) c[i]+=c[i-1];    for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];}void get_sa(char c[],int n,int sz=128){    x=rank,y=rank2;    for(int i=0;i<n;i++) x[i]=c[i],y[i]=i;    radix_sort(n,sz);    for(int len=1;len<n;len<<=1)    {        int yid=0;        for(int i=n-len;i<n;i++) y[yid++]=i;        for(int i=0;i<n;i++) if(sa[i]>=len) y[yid++]=sa[i]-len;        radix_sort(n,sz);        swap(x,y);        x[sa[0]]=yid=0;        for(int i=1;i<n;i++)        {            x[sa[i]]=cmp(y,sa[i],sa[i-1],len,n)?yid:++yid;        }        sz=yid+1;        if(sz>=n) break;    }    for(int i=0;i<n;i++) rank[i]=x[i];}void get_h(char str[],int n){    int k=0; h[0]=0x3f3f3f3f;    for(int i=0;i<n;i++)    {        if(rank[i]==0) continue;        k=max(k-1,0);        int j=sa[rank[i]-1];        while(i+k<n&&j+k<n&&str[i+k]==str[j+k]) k++;        h[rank[i]]=k;    }}int dp[maxn][20],Log[maxn];void RMQ_init(int n){    for(int i=0;i<n;i++) dp[i][0]=h[i];    for(int i=1;i<=Log[n];i++)    {        for(int j=0;j+(1<<i)-1<n;j++)        {            dp[j][i]=min(dp[j][i-1],dp[j+(1<<(i-1))][i-1]);        }    }}int lcp(int l,int r){    l=rank[l];r=rank[r];    if(l>r) swap(l,r);    int a=l+1,b=r;    int k=Log[b-a+1];    return min(dp[a][k],dp[b-(1<<k)+1][k]);}int main(){    int cas=1;    Log[0] = -1;for(int i=1;i<=maxn;i++)    {Log[i]=(i&(i-1))?Log[i-1]:Log[i-1] + 1 ;}    while(scanf("%s",str)!=EOF&&str[0]!='#')    {        int n=strlen(str);        get_sa(str,n);        get_h(str,n);        RMQ_init(n);        int l,r,t,k,maxx=-1,a;        for(l=1;l<n/2;l++)        {            for(int i=0;i+l<n;i+=l)            {                k=lcp(i,i+l);                r=k/l+1;                t=l-k%l;                t=i-t;                if(t>=0&&k%l!=0)                {                    if(lcp(t,t+l)>=k) r++;                }                if(r>maxx)                {                    a=0;                    maxx=r;                    ans[a++]=l;                }                else if(r==maxx)                {                    ans[a++]=l;                }            }        }        int pos;        int st,b=0,c;        for(int i=0;i<n&&!b;i++)        {            if(b) break;            for(int j=0;j<a;j++)            {                int tl=ans[j];                if(lcp(sa[i],sa[i]+tl)>=(maxx-1)*tl)                {                    st=sa[i];                    pos=i;                    l=tl*maxx;                    b=1;                    break;                }            }        }        printf("Case %d: ",cas++);        for (int i=0;i<l;i++) putchar(str[st+i]);        putchar(10);    }    return 0;}