code/leetcode/v1/[DP]MSM

矩阵连乘 (Multiplying a sequence of matrices)

我们想从两个以上矩阵连乘的式子中得到最佳的次序,使得计算时乘法运算最少。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
public class DynamicProgramming {

private static final int N = 5;

public static void main(String[] args) {
int[][] C = new int[N][N];
int[][] S = new int[N][N];
int[] r = {4, 5, 3, 6, 4, 5};

for (int i = 0; i < N; i++) {// 初始化 最小乘运算数组
for (int j = 0; j < N; j++) {
C[i][j] = 0;
}
}

for (int i = 0; i < N; i++) {// 初始化 分割位置
for (int j = 0; j < N; j++) {
S[i][j] = 0;
}
}

for (int p = 2; p <= N; p++) {// P 为间隔
for (int i = 0, j; i < N - p + 1; i++) {// i 为起始值
j = i + p - 1;// j 为终止值
C[i][j] = C[i][i] + C[i + 1][j] + r[i] * r[i + 1] * r[j + 1];// 赋 C[i][j] 的初值方便下文比较取得最小值
S[i][j] = i;
for (int k = i; k< j; k++) {// k 为中间分割值
int temp = C[i][k] + C[k + 1][j] + r[i] * r[k + 1] * r[j + 1];

if (temp < C[i][j]) {// 取得每次分割后的最小值
C[i][j] = temp;
S[i][j] = k;
}
}
}
}

for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
System.out.print(C[i][j] + "\t");
}
System.out.println();
}

System.out.println();

for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (j > i) System.out.print(S[i][j] + 1 + "\t");
else System.out.print(0 + "\t");
}
System.out.println();
}
}
}

/* result
0 60 132 180 252
0 0 90 132 207
0 0 0 72 132
0 0 0 0 120
0 0 0 0 0

0 1 2 2 2
0 0 2 2 2
0 0 0 3 4
0 0 0 0 4
0 0 0 0 0

(A1 * A2) * ((A3 * A4) * A5)
*/