POJ3368 Frequent values

Description

You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1
≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers ai , ... , aj.

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 , ... , an (-100000
≤ ai ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: ai ≤ ai+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.

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：

ª对于线段树上的节点k：
wif lSon[k]这一区间上所有的值相等并等于rSon[k]最左边的值 then  left[k] = left[lSon[k]] + left[rSon[k]]
     else  left[k] = left[lSon[k]]
wIf rSon[k]这一区间上所有的值相等并等于lSon[k]最右边的值 then  right[k] = right[rSon[k]] + right[lSon[k]]
     else  right[k] = right[rSon[k]]
wIf lSon[k]最右边的值与rSon[k]最左边的值相等 then
max[k] = Max{max[lSon[k]], max[rSon[k]], right[lSon[k]] + left[rSon[k]]}
else max[k] = Max{max[lSon[k]], max[rSon[k]]} 

代码：
if a[tree[p*2].r]=a[tree[p*2+1].l] then  

    begin  

      tree[p].max:=tree[p*2].r+tree[p*2+1].fl;  

      if a[tree[p].r]=a[tree[p*2+1].l] then  

        tree[p].r:=tree[p].max;  

      if a[tree[p].l]=a[tree[p*2].r] then  

        tree[p].l:=tree[p].max;  

    end else tree[p].max:=1;  

  if tree[p*2].max>tree[p].max then  

    tree[p].max:=tree[p*2].max;  

  if tree[p*2+1].max>tree[p].max then  

    tree[p].max:=tree[p*2+1].max;