PKU 3368 Frequent values 线段树

题意/Description：

  给一个长度为n的不降序列a1,a2,a3,…,an,有q个询问，每个询问为:

    i j

  询问在子序列ai…aj中出现最多的元素。

  数据范围：1 <= n, q <= 100000

读入/Input：

    The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a₁ , ... , a_n (-100000 ≤ a_i ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: a_i ≤ a_i+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the query.

    The last test case is followed by a line containing a single 0.

输出/Output：

    For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

题解/solution：

  注意到题目描述中的“不降序列”，让我们联想到可以使用线段树这一数据结构。

  在线段树的结点内设5个变量l、r、fmax、fl、fr，[l,r]表示该结点的区间范围，lf和rf分别表示元素a[l]和a[r]在区间内的出现频率，fmax表示区间内的最高出现频率。

  假设区间[x,y]和[y+1,z]均被询问[i,j]覆盖，则可以分情况讨论区间[x,z]的fmax值：

  若a[y]==a[y+1]，则fmax[x,y]=max{fmax[x,y],fmax[y+1,z],rf[x,y]+lf[y+1,z]}

  否则fmax[x,y]=max{fmax[x,y],fmax[y+1,z]}

代码/Code：

type  arr=record    l,r:longint;    fl,fr,fmax:longint;  end;var  tree:array [0..300001] of arr;  a:array [0..100001] of longint;  n,m:longint;function min(o,p:longint):longint;begin  if o<p then exit(o);  exit(p);end;procedure ins(p,b,e:longint);var  m:longint;begin  tree[p].l:=b; tree[p].r:=e;  if (b=e) then    begin      tree[p].fl:=1;      tree[p].fr:=1;      tree[p].fmax:=1;      exit;    end;  m:=(b+e) shr 1;  ins(p*2,b,m);  ins(p*2+1,m+1,e);  tree[p].fl:=tree[p*2].fl;  tree[p].fr:=tree[p*2+1].fr;  if a[tree[p*2].r]=a[tree[p*2+1].l] then    begin      tree[p].fmax:=tree[p*2].fr+tree[p*2+1].fl;      if a[tree[p].r]=a[tree[p*2+1].l] then        tree[p].fr:=tree[p].fmax;      if a[tree[p].l]=a[tree[p*2].r] then        tree[p].fl:=tree[p].fmax;    end else tree[p].fmax:=1;  if tree[p*2].fmax>tree[p].fmax then    tree[p].fmax:=tree[p*2].fmax;  if tree[p*2+1].fmax>tree[p].fmax then    tree[p].fmax:=tree[p*2+1].fmax;end;function count(p,b,e:longint):longint;var  max1,max2:longint;begin  with tree[p] do    begin      if (b<=l) and (e>=r) then exit(fmax);      if e<=tree[p*2].r then        exit(count(p*2,b,e));      if b>=tree[p*2+1].l then        exit(count(p*2+1,b,e));      count:=1;      if a[tree[p*2].r]=a[tree[p*2+1].l] then        count:=min(tree[p*2].r-b+1,tree[p*2].fr)+min(e-tree[p*2+1].l+1,tree[p*2+1].fl);  // 询问区间有可能未把tree[p*2].fr个元素全包含进去，所以要取min(tree[p*2].r-b+1, tree[p*2].fr)      max1:=count(p*2,b,tree[p*2].r);      max2:=count(p*2+1,tree[p*2+1].l,e);      if max1>count then count:=max1;      if max2>count then count:=max2;    end;end;procedure main;var  i,x,y:longint;begin  while 1=1 do    begin      read(n);      if n=0 then exit;      readln(m);      for i:=1 to n do        read(a[i]);      ins(1,1,n);      for i:=1 to m do        begin          readln(x,y);          writeln(count(1,x,y));        end;    end;end;begin  main;end.