poj3693

【题意】

给定一个字符串（长度不超过100000），要求输出重复次数最多的字典序最小的子串

【输入】

多组数据，每组数据一行，为一个小写字母组成的字符串

数据以#结束

【输出】

每组输出'Case *: '+重复次数最多字典序最小的子串

后缀数组求解

然后枚举重复部分的长度l

然后在l+1、2*l+1...位置枚举

若枚举到l+1若1和l+1上的字符相同，

则求1和l+1向前匹配的长度和向后匹配的长度。

然后将向前向后匹配长度相加再加上l然后div l

便是重复次数p

字典序的比较，比较一下起点的rank即可

需要注意的是向前匹配的最左端到最右端向前p*l都可以作为字符串的起点

用st算法可求出rank最小的起点

program poj3693;const  ln2=1/ln(2);type  arr=array [-1..100001] of longint;  arra=array [0..18,0..100001] of longint;var  n,i,j,k,l,max,where,t,s,e,mid,long,q,p,tot:longint;  ok:boolean;  root,nroot:ansistring;  nrank,nheight,sa,dl,height,total,rank,now,keep,change:arr;  min,dict,nmin:arra;procedure swap (var a,b:longint);var  i:longint;begin  i:=a;  a:=b;  b:=i;end;procedure solve (var rank,height:arr;root:ansistring);begin  fillchar(total,sizeof(total),0);  for i:=1 to n do    inc(total[ord(root[i])-96]);  for i:=1 to 26 do    total[i]:=total[i-1]+total[i];  for i:=1 to n do    rank[i]:=total[ord(root[i])-97]+1;  k:=0;  while 1 shl k<n do    begin      for i:=1 to n do        if i+(1 shl k)>n then change[i]:=0                         else change[i]:=rank[i+(1 shl k)];      fillchar(total,sizeof(total),0);      for i:=1 to n do        inc(total[change[i]]);      for i:=1 to n do        total[i]:=total[i]+total[i-1];      for i:=1 to n do        begin          dl[total[change[i]-1]+1]:=i;          inc(total[change[i]-1]);        end;      fillchar(total,sizeof(total),0);      ok:=true;      for i:=1 to n do        begin          inc(total[rank[i]]);          if total[rank[i]]=2 then ok:=false;        end;      if ok then break;      for i:=2 to n do        total[i]:=total[i]+total[i-1];      fillchar(now,sizeof(now),0);      fillchar(keep,sizeof(keep),0);      for i:=1 to n do        begin          if now[rank[dl[i]]]<>change[dl[i]] then            begin              now[rank[dl[i]]]:=change[dl[i]];              total[rank[dl[i]]-1]:=total[rank[dl[i]]-1]+keep[rank[dl[i]]];              keep[rank[dl[i]]]:=0;            end;          inc(keep[rank[dl[i]]]);          rank[dl[i]]:=total[rank[dl[i]]-1]+1;        end;      inc(k);    end;  for i:=1 to n do    sa[rank[i]]:=i;  fillchar(height,sizeof(height),0);  for i:=1 to n do    begin      if rank[i]=1 then continue;      k:=height[rank[i-1]]-1;      if k<0 then k:=0;      while (i+k<=n)and(sa[rank[i]-1]+k<=n)and(root[i+k]=root[sa[rank[i]-1]+k]) do inc(k);      height[rank[i]]:=k;    end;end;procedure st (var height:arr;var min:arra);begin  for i:=1 to n do    min[0,i]:=height[i];  k:=0;  while 1 shl k<n do    begin      for i:=1 to n do        if i+(1 shl k)>n then          min[k+1,i]:=min[k,i]                         else        if min[k,i]<min[k,i+(1 shl k)] then          min[k+1,i]:=min[k,i]                                       else          min[k+1,i]:=min[k,i+(1 shl k)];      inc(k);    end;end;procedure dictst;begin  for i:=1 to n do    dict[0,i]:=i;  k:=0;  while 1 shl k<n do    begin      for i:=1 to n do        if i+(1 shl k)>n then          dict[k+1,i]:=dict[k,i]                         else        if rank[dict[k,i]]<rank[dict[k,i+(1 shl k)]] then          dict[k+1,i]:=dict[k,i]                                                     else          dict[k+1,i]:=dict[k,i+(1 shl k)];      inc(k);    end;end;function similar (a,b:longint;var min:arra;var rank,height:arr):longint;var  i:longint;begin  if a=b then exit(n-a+1);  a:=rank[a];  b:=rank[b];  if a>b then swap(a,b);  inc(a);  if a=b then exit(height[a]);  i:=trunc(ln(b-a)*ln2);  while (i>0)and(1 shl i>=b-a+1) do dec(i);  if min[i,a]<min[i,b-(1 shl i)+1] then exit(min[i,a])                                   else exit(min[i,b-(1 shl i)+1]);end;function similardict (a,b:longint):longint;var  i:longint;begin  if a=b then exit(a);  i:=trunc(ln(b-a)*ln2);  while (i>0)and(1 shl i>=b-a+1) do dec(i);  if rank[dict[i,a]]<rank[dict[i,b-(1 shl i)+1]] then exit(dict[i,a])                                                 else exit(dict[i,b-(1 shl i)+1]);end;begin  t:=0;  repeat    readln(root);    if root='#' then break;    inc(t);    n:=length(root);    tot:=0;    solve(rank,height,root);    st(height,min);    nroot:=root;    for i:=1 to n do      nroot[i]:=root[n-i+1];    solve(nrank,nheight,nroot);    st(nheight,nmin);    dictst;    max:=1;    long:=1;    where:=1;    for i:=2 to n do      if root[i]<root[where] then where:=i;    for l:=1 to n div 2 do      if n div l >= max then        for i:=1 to n div l  do          if n div l - i + 3>=max then          begin            if i*l+1>n then continue;            if root[(i-1)*l+1]<>root[i*l+1] then continue;            s:=(i-1)*l+1-similar(n-(i-1)*l+1,n-i*l+1,nmin,nrank,nheight);            k:=similar(s,s+l,min,rank,height)+l;            q:=k div l;            if q>=max then              begin                p:=similardict(s,s+k-q*l);                if (q>max)or                ((q=max)and(rank[where]>rank[p])) then                  begin                    max:=q;                    where:=p;                    long:=l;                  end;              end;          end;    write('Case ',t,': ');    for i:=where to where+max*long-1 do      write(root[i]);    writeln;  until false;end.