mirror of
https://github.com/SunnyQjm/algorithm-review.git
synced 2026-06-03 08:16:43 +08:00
54 lines
2.0 KiB
Python
54 lines
2.0 KiB
Python
#!/usr/bin/env python
|
||
# coding=utf-8
|
||
|
||
#######################################################################################
|
||
# 矩阵相乘 => 详情见PPT
|
||
#
|
||
# 描述：设 A1, A2, … , An 为矩阵序列，Ai 为Pi-1xPi 阶矩阵，i = 1,2,…,n. 确定乘法顺序使得元素相乘的总次数最少.
|
||
# 输入：向量 P = <P0, P1, … , Pn>，n个矩阵的行数、列数
|
||
#
|
||
# 实例：P = <10, 100, 5, 50>
|
||
# A1: 10x100, A2: 100x5, A3: 5x50
|
||
# 括号位置不同，相乘总次数不同：
|
||
# (A1 A2)A3: 10x100x5 + 10x5x50 = 7500
|
||
# A1(A2 A3): 10x100x50 + 100x5x50 = 75000
|
||
#######################################################################################
|
||
|
||
class Solution:
|
||
def miniMultiplicationTimes(self, matrixs):
|
||
"""
|
||
:type matrixs: List[int]
|
||
:rtype int
|
||
|
||
(knowledge)
|
||
|
||
思路：
|
||
1. 使用动态规划；
|
||
2. dp[i][j] => 得到Ai..j 所需要的最少相乘次数
|
||
3. 状态转移方程：
|
||
m(i, j) = 0 i <= j <= i + 1
|
||
min(i < k < j){m(i, k) + m(k, j) + Pi*Pk*Pj} j > i + 1
|
||
"""
|
||
|
||
# 初始化dp数组
|
||
dp = [[float("inf") for i in range(len(matrixs))] for i in range(len(matrixs))]
|
||
|
||
# 让满足 i <= j <= i + 1的置为0
|
||
for i in range(len(matrixs) - 1):
|
||
dp[i][i] = 0
|
||
dp[i][i + 1] = 0
|
||
dp[len(matrixs) - 1][len(matrixs) - 1] = 0
|
||
|
||
# 此处i表示每轮要求的目标值的间隔
|
||
# 比如i=2，表示求所有索引间隔为2得dp值，eg.: dp[0][2], dp[1][3], dp[2][4]...
|
||
for i in range(2, len(matrixs)):
|
||
for j in range(len(matrixs) - i):
|
||
for k in range(j + 1, j + i):
|
||
dp[j][j + i] = min(dp[j][j + i], dp[j][k] + dp[k][j + i] + matrixs[j] * matrixs[k] * matrixs[j + i])
|
||
return dp[0][len(matrixs) - 1]
|
||
|
||
|
||
if __name__ == '__main__':
|
||
solution = Solution()
|
||
print(solution.miniMultiplicationTimes([30, 35, 15, 5, 10, 20]), "= 11875")
|