poj2452

【题意】

有n个数序列s，求出满足s[i]<s[k]<s[j](i<k<j,j>=i+1)的最大j-i的值

【输入】

多组数据

每组数据第一行为n

第二行为n个数，描述s

【输出】

对于每组数据输出最大的j-i的值

可以转化为rmq问题

枚举i，可以通过二分找到满足之间最小数大于p[i]的最右端i`

然后又可以找到i+1到i`之间的最大值k

然后再通过二分查找最靠左的值为k的位置

program poj2452;type  arr=array [0..16,0..50001] of longint;var  n,i,j,k,s,e,ans,mid:longint;  p:array [0..50001] of longint;  max,min:arr;function wmax (a,b:longint;var max:arr;x:longint):longint;var  k:longint;begin  if a=b then exit(p[a]);  k:=trunc(ln(b-a+1)/ln(2));  if max[k,a]*x>max[k,b-(1 shl k)+1]*x then exit(max[k,a])                                       else exit(max[k,b-(1 shl k)+1]);end;procedure st (var max:arr;x:longint);begin  for i:=1 to n do    max[0,i]:=p[i];  k:=0;  while 1 shl k<n do    begin      for i:=1 to n do        if i+(1 shl k)>n then max[k+1,i]:=max[k,i]                         else        if max[k,i]*x>max[k,i+(1 shl k)]*x then max[k+1,i]:=max[k,i]                                           else max[k+1,i]:=max[k,i+(1 shl k)];      inc(k);    end;end;begin  while not seekeof do    begin      read(n);      for i:=1 to n do        read(p[i]);      if n<=1 then        begin          writeln(-1);          continue;        end;      st(max,1);      st(min,-1);      ans:=0;      for i:=1 to n-1 do        begin          if p[i+1]<=p[i] then continue;          s:=i+1;          e:=n;          while s<>e do            begin              mid:=s+e-(s+e) div 2;              if wmax(i+1,mid,min,-1)>p[i] then s:=mid                                           else e:=mid-1;            end;          k:=wmax(i+1,s,max,1);          if p[i+1]=k then            begin              if ans<1 then ans:=1;              continue;            end;          s:=i+2;          while s<>e do            begin              mid:=s+e-(s+e) div 2;              if wmax(i+1,mid-1,max,1)<k then s:=mid                                         else e:=mid-1;            end;          if e-i>ans then ans:=e-i;        end;      if ans=0 then writeln(-1)               else writeln(ans);    end;end.