DDR爱好者之家 Design By 杰米

算法导论上的伪码改写而成,加上导论的课后练习第一题的解的构造函数。
复制代码 代码如下:
#encoding: utf-8
=begin
author: xu jin
date: Nov 11, 2012
Optimal Binary Search Tree
to find by using EditDistance algorithm
refer to <<introduction to algorithms
example output:
"k2 is the root of the tree."
"k1 is the left child of k2."
"d0 is the left child of k1."
"d1 is the right child of k1."
"k5 is the right child of k2."
"k4 is the left child of k5."
"k3 is the left child of k4."
"d2 is the left child of k3."
"d3 is the right child of k3."
"d4 is the right child of k4."
"d5 is the right child of k5."

The expected cost is 2.75. 
=end

INFINTIY = 1 / 0.0
a = ['', 'k1', 'k2', 'k3', 'k4', 'k5']
p = [0, 0.15, 0.10, 0.05, 0.10, 0.20]
q = [0.05, 0.10, 0.05, 0.05, 0.05 ,0.10]
e = Array.new(a.size + 1){Array.new(a.size + 1)}
root = Array.new(a.size + 1){Array.new(a.size + 1)}

def optimalBST(p, q, n, e, root)
  w = Array.new(p.size + 1){Array.new(p.size + 1)}
  for i in (1..n + 1)
    e[i][i - 1] = q[i - 1]
    w[i][i - 1] = q[i - 1]
  end
  for l in (1..n)
    for i in (1..n - l + 1)
      j = i + l -1
      e[i][j] = 1 / 0.0
      w[i][j] = w[i][j - 1] + p[j] + q[j]
      for r in (i..j)
        t = e[i][r - 1] + e[r + 1][j] + w[i][j]
        if t < e[i][j]
          e[i][j] = t
          root[i][j] = r
        end
      end
    end
  end
end

def printBST(root, i ,j, signal)
  return if i > j
  if signal == 0
   p "k#{root[i][j]} is the root of the tree."
   signal = 1
  end
  r = root[i][j]
  #left child
  if r - 1< i
    p "d#{r - 1} is the left child of k#{r}."
  else
    p "k#{root[i][r - 1]} is the left child of k#{r}."
    printBST(root, i, r - 1, 1 )
  end
  #right child
  if r >= j
     p "d#{r} is the right child of k#{r}."
  else
    p "k#{root[r + 1][j]} is the right child of k#{r}."
    printBST(root, r + 1, j, 1)
  end
 
end

optimalBST(p, q, p.size - 1, e, root)
printBST(root, 1, a.size-1, 0)
puts "\nThe expected cost is #{e[1][a.size-1]}."

DDR爱好者之家 Design By 杰米
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DDR爱好者之家 Design By 杰米

RTX 5090要首发 性能要翻倍!三星展示GDDR7显存

三星在GTC上展示了专为下一代游戏GPU设计的GDDR7内存。

首次推出的GDDR7内存模块密度为16GB,每个模块容量为2GB。其速度预设为32 Gbps(PAM3),但也可以降至28 Gbps,以提高产量和初始阶段的整体性能和成本效益。

据三星表示,GDDR7内存的能效将提高20%,同时工作电压仅为1.1V,低于标准的1.2V。通过采用更新的封装材料和优化的电路设计,使得在高速运行时的发热量降低,GDDR7的热阻比GDDR6降低了70%。