【BZOJ1717&POJ3261】Milk Patterns（后缀数组，二分）

题意：求字符串的可重叠的k次最长重复子串

n<=20000 a[i]<=1000000

思路：后缀数组+二分答案x，根据height分组，每组之间的height>=x

因为可以重叠，所以只要判断是否有一组的height个数>=k即可

 var sa,rank,x,y,a,wc,wd,height:array[..]of longint;
     n,m,i,l,r,mid,last,k1:longint;

 procedure swap(var x,y:longint);
 var t:longint;
 begin
  t:=x; x:=y; y:=t;
 end;

 function cmp(a,b,l:longint):boolean;
 begin
  exit((y[a]=y[b])and(y[a+l]=y[b+l]));
 end;

 procedure getsa(n:longint);
 var i,j,p:longint;
 begin
  for i:= to n- do
  begin
   x[i]:=a[i];
   inc(wc[a[i]]);
  end;
  for i:= to m- do wc[i]:=wc[i-]+wc[i];
  for i:=n- downto  do
  begin
   dec(wc[x[i]]);
   sa[wc[x[i]]]:=i;
  end;
  j:=; p:=;
  while p<n do
  begin
   p:=;
   for i:=n-j to n- do
   begin
    y[p]:=i; inc(p);
   end;
   for i:= to n- do
    if sa[i]>=j then begin y[p]:=sa[i]-j; inc(p); end;
   for i:= to n- do wd[i]:=x[y[i]];
   for i:= to m- do wc[i]:=;
   for i:= to n- do inc(wc[wd[i]]);
   for i:= to m- do wc[i]:=wc[i-]+wc[i];
   for i:=n- downto  do
   begin
    dec(wc[wd[i]]);
    sa[wc[wd[i]]]:=y[i];
   end;
   for i:= to n do swap(x[i],y[i]);
   p:=; x[sa[]]:=;
   for i:= to n- do
    if cmp(sa[i-],sa[i],j) then x[sa[i]]:=p-
     else begin x[sa[i]]:=p; inc(p); end;
   j:=j*;
   m:=p;
  end;
 end;

 procedure getheight(n:longint);
 var i,k,j:longint;
 begin
  for i:= to n do rank[sa[i]]:=i;
  k:=;
  for i:= to n- do
  begin
   if k> then dec(k);
   j:=sa[rank[i]-];
   while a[i+k]=a[j+k] do inc(k);
   height[rank[i]]:=k;
  end;
 end;

 function isok(x:longint):boolean;
 var s,i:longint;
 begin
  s:=;
  for i:= to n do
  begin
   if height[i]<x then
   begin
    if s>=k1 then exit(true);
    s:=;
   end
    else inc(s);
  end;
  if s>=k1 then exit(true);
  exit(false);
 end;

 begin

  while not eof do
  begin
   fillchar(a,sizeof(a),);
   fillchar(sa,sizeof(sa),);
   fillchar(rank,sizeof(rank),);
   fillchar(x,sizeof(x),);
   fillchar(y,sizeof(y),);
   fillchar(height,sizeof(height),);
   fillchar(wc,sizeof(wc),);
   fillchar(wd,sizeof(wd),);
   readln(n,k1);
   if n= then break;
   for i:= to n- do
   begin
    read(a[i]);
    inc(a[i]);
   end;
   a[n]:=; m:=;
   getsa(n+);
   getheight(n);
   l:=; r:=n; last:=;
   while l<=r do
   begin
    mid:=(l+r)>>;
    if isok(mid) then begin last:=mid; l:=mid+; end
     else r:=mid-;
   end;
   writeln(last);
  // for i:= to n do writeln(height[i]);
  end;

 end.