bzoj3790

观察发现，这道题目其实就相当于一个最小区间覆盖问题
这里的区间是指以每个点为中心的最长回文串
很久没写manacher，有点感动
不得不说manacher是一个非常好的算法

 var s:array[..] of char;
     c,l,r,f,p:array[..] of longint;
     i,j,t,n,m,ans,k,right:longint;
     ch:char;

 procedure swap(var a,b:longint);
   var c:longint;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

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

 procedure sort(ll,rr: longint);
   var i,j,x,y: longint;
   begin
     i:=ll;
     j:=rr;
     x:=r[(ll+rr) shr ];
     y:=l[(ll+rr) shr ];
     repeat
       while (r[i]<x) or (r[i]=x) and (l[i]<y) do inc(i);
       while (x<r[j]) or (r[j]=x) and (y<l[j]) do dec(j);
       if not(i>j) then
       begin
         swap(r[i],r[j]);
         swap(l[i],l[j]);
         inc(i);
         j:=j-;
       end;
     until i>j;
     if ll<j then sort(ll,j);
     if i<rr then sort(i,rr);
   end;

 function ask(x:longint):longint;
   begin
     if x= then exit();
     ask:=n+;
     while x<=n do  //因为求的是后缀最小值，事实上只要将查询和修改的范围换一下即可
     begin
       ask:=min(ask,c[x]);
       x:=x+lowbit(x);
     end;
   end;

 procedure add(i:longint);
   var x:longint;
   begin
     x:=r[i];
     while x> do
     begin
       c[x]:=min(c[x],f[i]);
       x:=x-lowbit(x);
     end;
   end;

 begin
   while true do
   begin
     read(ch);
     if not((ch>='a') and (ch<='z')) then break;
     s[]:='$';
     s[]:='#';
     m:=;
     n:=;
     while (ch>='a') and (ch<='z') do
     begin
       inc(n);
       inc(m);
       s[m]:=ch;
       inc(m);
       s[m]:='#'; //将回文串的奇偶转化为一种情况
       read(ch);
     end;
     s[m+]:=' ';
     readln;
     k:=;
     right:=;
     t:=;
     fillchar(p,sizeof(p),);
     for i:= to m do
     begin
       if right>i then   //manacher的核心
         p[i]:=min(p[*k-i],right-i)
       else p[i]:=;
       while s[i+p[i]]=s[i-p[i]] do inc(p[i]);
       if p[i]+i>right then
       begin
         right:=p[i]+i;
         k:=i;
       end;
       if (s[i]='#') and (p[i]=) then continue
       else begin
         inc(t);
         l[t]:=(i-p[i]+) div ;
         r[t]:=(i+p[i]-) div ;
       end;
     end;
     sort(,t);  //解决区间覆盖问题，右端点排序
     ans:=n;
     for i:= to n do
       c[i]:=n+;
     for i:= to t do
     begin
       f[i]:=ask(l[i]-)+; //以前这个问题我都是用单调队列的，这里写了树状数组，更简明
       if r[i]=n then ans:=min(ans,f[i]);
       add(i);
     end;
     writeln(ans-);
   end;
 end.