技術標籤：python演算法-每日一練演算法python資料結構java線性代數

演算法實驗報告2——矩陣連乘問題

• • 一丶矩陣連乘問題
  • 二丶矩陣連乘程式碼
  • 三丶執行結果
  • • • • • ——虎鼓瑟兮鸞回車，仙之人兮列如麻——

一丶矩陣連乘問題

A1 A2 A3 A4 A5 A6
3035 3515 155 510 1020 2025
求最小數乘次數。

二丶矩陣連乘程式碼

# coding=gbk
__author__ = 'Blockchain_Key'
__time__ = '2020.12.17'

# row_num 行數
# col_num 列數
class Matrix:
    def
 
 __init__(self, row_num=0, col_num=0, matrix=None):
        if matrix != None:
            self.row_num = len(matrix)
            self.col_num = len(matrix[0])
        else:
            self.row_num = row_num
            self.col_num = col_num
        self.matrix = matrix


def matrix_chain(matrixs):
    matrix_num =
 
 len(matrixs)     # 求取矩陣個數
    count = [[0 for j in range(matrix_num)] for i in range(matrix_num)]   # 建立二維動態規劃表
    flag = [[0 for j in range(matrix_num)] for i in range(matrix_num)]  # 建立二維標識矩陣
    for interval in range(1, matrix_num + 1):   # 從1開始，方便計算
        for i in range(matrix_num - interval):
 
  # 剩餘的矩陣
            j = i + interval
            count[i][j] = count[i][i] + count[i + 1][j] + matrixs[i].row_num * matrixs[i + 1].row_num * matrixs[j].col_num
            flag[i][j] = i
            for k in range(i + 1, j):
                temp = count[i][k] + count[k + 1][j] + matrixs[i].row_num * matrixs[k + 1].row_num * matrixs[j].col_num
                if temp < count[i][j]:
                    count[i][j] = temp
                    flag[i][j] = k
    traceback(0, matrix_num - 1, flag)
    return count[0][matrix_num - 1]  # 最終結果


def traceback(i, j, flag):
    if i == j:
        return
    # 遞迴處理
    traceback(i, flag[i][j], flag)
    traceback(flag[i][j] + 1, j, flag)


if __name__=="__main__":
    matrixs = [Matrix(30, 35), Matrix(35, 15), Matrix(15, 5), Matrix(5, 10), Matrix(10, 20), Matrix(20, 25)]
    result = matrix_chain(matrixs)
    print(result)

三丶執行結果

——虎鼓瑟兮鸞回車，仙之人兮列如麻——