【BZOJ3790】神奇项链（manacher，树状数组）

题意：

思路：生成一些回文拼起来使生成的段数最小

显然存在一种最优的方案，使生成的那些回文是目标串的极长回文子串

求出对于每个位置的最长回文子串，问题就转化成了：

给定一些已知起始和终止位置的线段，求覆盖住整个区域的最小线段数量

这个可以BIT做，求当前已经覆盖的区域最远能拓展到哪里

也可以预处理一下前缀最小值，跳转时直接调用即可

const oo=1000000000;
var t,a,x,y,p:array[1..110000]of longint;
    len,n,i,id,mx,ans,m:longint;
    ch:ansistring;

function lowbit(x:longint):longint;
begin
 exit(x and (-x));
end;

function max(x,y:longint):longint;
begin
 if x>y then exit(x);
 exit(y);
end;

function min(x,y:longint):longint;
begin
 if x<y then exit(x);
 exit(y);
end;

procedure update(x,y:longint);
begin
 while x<=n-2 do
 begin
  t[x]:=max(t[x],y);
  x:=x+lowbit(x);
 end;
end;

function query(x:longint):longint;
begin
 query:=-oo;
 while x>0 do
 begin
  query:=max(query,t[x]);
  x:=x-lowbit(x);
 end;
end;

begin
 assign(input,'bzoj3790.in'); reset(input);
 assign(output,'bzoj3790.out'); rewrite(output);
 while not eof do
 begin
  readln(ch);
  len:=length(ch);
  if len=0 then break;
  fillchar(a,sizeof(a),0);
  fillchar(p,sizeof(p),0);
  n:=2; a[1]:=27; a[2]:=28;
  for i:=1 to len do
  begin
   inc(n); a[n]:=ord(ch[i])-ord('a')+1;
   inc(n); a[n]:=28;
  end;
  inc(n); a[n]:=29;
  mx:=0; id:=1;
  for i:=2 to n-1 do
  begin
   if mx>i then p[i]:=min(p[id*2-i],mx-i)
    else p[i]:=1;
   while a[i-p[i]]=a[i+p[i]] do inc(p[i]);
   if p[i]+i>mx then
   begin
    mx:=p[i]+i; id:=i;
   end;
  end;
  for i:=1 to m do
  begin
   x[i]:=0; y[i]:=0;
  end;
  m:=0;
  for i:=2 to n-1 do
  begin
   inc(m); x[m]:=i-p[i]; y[m]:=i+p[i]-2;
  end;
  fillchar(t,sizeof(t),0);
  for i:=1 to m do update(x[i],y[i]);
  i:=1; ans:=0;
  while i<n-2 do
  begin
   i:=query(i+1);
   inc(ans);
  end;
  writeln(ans-1);
 end;
 close(input);
 close(output);
end.