import numpy as np def MatrixChain(p,n): #存储最优值 m = np.zeros((n+1,n+1)) # 牺牲第0行，第0列 s = np.zeros((n+1,n+1)) # 牺牲第0行，第0列 # 单个矩阵连乘的次数 for i in range(n+1): #对角线值为0 m[i][i] = 0; s[i][i] = 0; # r表示子问题的规模，即连乘的矩阵个数，从两个矩阵开始，规模逐步放大到n for r in range(2,n+1): for i in range(1,n-r+2): j =i+r-1 # 当前的子问题 m[i][j] =m[i+1][j]+p[i-1]*p[i]*p[j] s[i][j] =i for k in range(i+1,j): t = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j] if t<m[i][j]: m[i][j] = t s[i][j] = k return m,s # 备忘录 def Traceback(i,j,s): global res # 全局变量 if i == j: res.append('A'+ str(i)) else: res.append('(') Traceback(i,int(s[i][j]),s) Traceback(int(s[i][j])+1,j,s) res.append(')') if __name__ == '__main__': arr = [] for i in range(5): row = input(f"请输入第{i + 1}行，用空格分隔每个元素：") arr.append(row.split(" ")) n = int(len(arr)) res = [] # 处理矩阵的行和列 p = [] for i in range(n): if i==0: p.append(arr[0][0]) p.append(arr[0][1]) else: p.append(arr[i][1]) m,s = MatrixChain(p,n) Traceback(1,n,s) print(''.join(res))修改这段代码

以下是修改后的代码，主要是将输入方式改为了直接输入数字，同时纠正了一些语法错误：

import numpy as np 

def MatrixChain(p, n):
    # 存储最优值
    m = np.zeros((n+1, n+1))  # 牺牲第0行，第0列
    s = np.zeros((n+1, n+1))  # 牺牲第0行，第0列
    
    # 单个矩阵连乘的次数
    for i in range(n+1):  # 对角线值为0
        m[i][i] = 0
        s[i][i] = 0
    
    # r表示子问题的规模，即连乘的矩阵个数，从两个矩阵开始，规模逐步放大到n
    for r in range(2, n+1):
        for i in range(1, n-r+2):
            j = i+r-1  # 当前的子问题
            m[i][j] = m[i+1][j] + p[i-1]*p[i]*p[j]
            s[i][j] = i
            for k in range(i+1, j):
                t = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j]
                if t < m[i][j]:
                    m[i][j] = t
                    s[i][j] = k
                    
    return m, s

# 备忘录
def Traceback(i, j, s):
    global res  # 全局变量
    if i == j:
        res.append('A'+str(i))
    else:
        res.append('(')
        Traceback(i, int(s[i][j]), s)
        Traceback(int(s[i][j])+1, j, s)
        res.append(')')

if __name__ == '__main__':
    arr = []
    for i in range(5):
        row = input(f"请输入第{i+1}行，用空格分隔每个元素：")
        arr.append(list(map(int, row.split())))
    n = len(arr)
    res = []
    
    # 处理矩阵的行和列
    p = []
    for i in range(n):
        if i == 0:
            p.append(arr[0][0])
            p.append(arr[0][1])
        else:
            p.append(arr[i][1])
            
    m, s = MatrixChain(p, n)
    Traceback(1, n, s)
    print(''.join(res))