Codeforces Round #361 (Div. 2) D Friends and Subsequences

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原题网址：http://codeforces.com/contest/689/problem/D

这里写图片描述

对于每一个确定的l,随着r的增大，max ai 单调增，min bi 单调减，对于每一个l可以二分，对于区间最值可以RMQ，总复杂度nlogn。（一开始区间最值用线段树平添log）

var  pr1,pr2:array[0..200050,0..18] of longint;  n,i,j,l,r,m:longint;  ans,t1,t2:int64;function max(a,b:longint):longint;  begin    if a>b      then exit(a)      else exit(b);  end;function min(a,b:longint):longint;  begin    if a<b      then exit(a)      else exit(b);  end;function f(x:longint):longint;  var    s:longint;  begin    s:=0;    while x>1 do      begin        x:=x>>1;        inc(s);      end;    exit(s);  end;function qmax(l,r:longint):longint;  var    t:longint;  begin    t:=f(r-l+1);    exit(max(pr1[l][t],pr1[r-1<<t+1][t]));  end;function qmin(l,r:longint):longint;  var    t:longint;  begin    t:=f(r-l+1);    exit(min(pr2[l][t],pr2[r-1<<t+1][t]));  end;begin  read(n);  for i:=1 to n do read(pr1[i][0]);  for i:=1 to n do read(pr2[i][0]);  for j:=1 to 18 do    for i:=1 to n do      begin        if i+1<<j-1>n then break;        pr1[i][j]:=max(pr1[i][j-1],pr1[i+1<<(j-1)][j-1]);        pr2[i][j]:=min(pr2[i][j-1],pr2[i+1<<(j-1)][j-1]);      end;  ans:=0;   for i:=1 to n do    begin         l:=i;r:=n;      while l<r do        begin          m:=(l+r+1)>>1;          if qmax(i,m)>qmin(i,m)            then r:=m-1            else l:=m;        end;      t1:=l;            l:=i;r:=n;      while l<r do        begin          m:=(l+r)>>1;          if qmax(i,m)<qmin(i,m)            then l:=m+1            else r:=m;        end;      t2:=l;            if (t1>=t2)and(qmax(i,t1)=qmin(i,t1))and(qmax(i,t2)=qmin(i,t2))        then inc(ans,t1-t2+1);          end;      writeln(ans);end.

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