Problem:
Score : 500 points

Problem Statement

We have N camels numbered 1,2,\ldots,N.
Snuke has decided to make them line up in a row.

The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.

Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.

Solve this problem for each of the T test cases given.

Constraints

All values in input are integers.

1 \leq T \leq 10^5

1 \leq N \leq 2 \times 10^{5}

1 \leq K_i \leq N

1 \leq L_i, R_i \leq 10^9

The sum of values of N in each input file is at most 2 \times 10^5.

Input

Input is given from Standard Input in the following format:

T
\mathrm{case}_1
\vdots
\mathrm{case}_T

Each case is given in the following format:

N
K_1 L_1 R_1
\vdots
K_N L_N R_N

Output

Print T lines. The i-th line should contain the answer to the i-th test case.

Sample Input 1

3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34

Sample Output 1

25
221
1354

In the first test case, it is optimal to line up the camels in the order 2, 1.

Camel 1 is not the frontmost camel, so its happiness will be 10.

Camel 2 is among the two frontmost camels, so its happiness will be 15.

In the second test case, it is optimal to line up the camels in the order 2, 1, 3.

Camel 1 is among the two frontmost camels, so its happiness will be 93.

Camel 2 is the frontmost camel, so its happiness will be 71.

Camel 3 is among the three frontmost camels, so its happiness will be 57.

Solution in Ruby (2.7.1) (judge verdict: Accepted)

class MyHeapQ
  def initialize(array = [])
    @data = []
    array.each{|a| push(a)}
  end

  def push(element)
    @data.push(element)
    bottom_up
  end

  def pop
    if size == 0
      return nil
    elsif size == 1
      return @data.pop
    else
      top = @data[0]
      @data[0] = @data.pop
      top_down
      return top
    end
  end

  def size
    @data.size
  end

  def data
    @data
  end

  private

  def swap(i, j)
    @data[i], @data[j] = @data[j], @data[i]
  end

  def parent_idx(target_idx)
    (target_idx - 1) / 2
  end

  def bottom_up
    target_idx = size - 1
    return if target_idx == 0
    parent_idx = parent_idx(target_idx)
    while ((@data[parent_idx][1] <=> @data[target_idx][1])>0)
      swap(parent_idx, target_idx)
      break if parent_idx == 0
      target_idx = parent_idx
      parent_idx = parent_idx(target_idx)
    end
  end

  def top_down
    target_idx = 0

    # child がある場合
    while (target_idx * 2 + 1 < @data.size)	# has_child?

      a = target_idx * 2 + 1		# left_child_idx
      b = target_idx * 2 + 2		# right_child_idx

      if @data[b].nil?
        c = a
      else
        c = (@data[a][1] <=> @data[b][1])<1 ? a : b
      end

      if (@data[target_idx][1] <=> @data[c][1])>0
        swap(target_idx, c)
        target_idx = c
      else
        return
      end
    end
  end
end

t = gets.strip.to_i
t.times do
  n = gets.strip.to_i
  klr = []
  s = []
  m = []
  ans = 0
  n.times do |i|
    k,l,r = gets.strip.split.map(&:to_i)
    if l == r
      ans += l
    elsif l > r
      s.push [k, l-r]
      ans += r
    else
      m.push [n-k, r-l]
      ans += l
    end
  end
  s.sort!
  m.sort!
  q = MyHeapQ.new
  c = 1
  s.each do |e|
    if e[0] > c
      while q.size > c
        q.pop
      end
      c = e[0]
    end
    q.push e
  end
  while q.size > c
    q.pop
  end
  while q.size > 0
    ans += q.data.pop[1]
  end
  c = 0
  m.each do |e|
    if e[0] > c
      while q.size > c
        q.pop
      end
      c = e[0]
    end
    q.push e
  end
  while q.size > c
    q.pop
  end
  while q.size > 0
    ans += q.data.pop[1]
  end
  p ans
end