Strassen的矩阵乘法方法是一种典型的分而治之算法。我们在这里讨论了Strassen算法。但是，让我们再次了解分而治之方法的背后并实现它。
先决条件：在进一步理解之前，必须先看这篇文章。
执行
// CPP program to implement Strassen’s Matrix
// Multiplication Algorithm
#include
using namespace std;
typedef long long lld;
/* Strassen's Algorithm for matrix multiplication
Complexity: O(n^2.808) */
inline lld** MatrixMultiply(lld** a, lld** b, int n,
int l, int m)
{
lld** c = new lld*[n];
for (int i = 0; i < n; i++)
c[i] = new lld[m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
c[i][j] = 0;
for (int k = 0; k < l; k++) {
c[i][j] += a[i][k] * b[k][j];
}
}
}
return c;
}
inline lld** Strassen(lld** a, lld** b, int n,
int l, int m)
{
if (n == 1 || l == 1 || m == 1)
return MatrixMultiply(a, b, n, l, m);
lld** c = new lld*[n];
for (int i = 0; i < n; i++)
c[i] = new lld[m];
int adjN = (n >> 1) + (n & 1);
int adjL = (l >> 1) + (l & 1);
int adjM = (m >> 1) + (m & 1);
lld**** As = new lld***[2];
for (int x = 0; x < 2; x++) {
As[x] = new lld**[2];
for (int y = 0; y < 2; y++) {
As[x][y] = new lld*[adjN];
for (int i = 0; i < adjN; i++) {
As[x][y][i] = new lld[adjL];
for (int j = 0; j < adjL; j++) {
int I = i + (x & 1) * adjN;
int J = j + (y & 1) * adjL;
As[x][y][i][j] = (I < n && J < l) ? a[I][J] : 0;
}
}
}
}
lld**** Bs = new lld***[2];
for (int x = 0; x < 2; x++) {
Bs[x] = new lld**[2];
for (int y = 0; y < 2; y++) {
Bs[x][y] = new lld*[adjN];
for (int i = 0; i < adjL; i++) {
Bs[x][y][i] = new lld[adjM];
for (int j = 0; j < adjM; j++) {
int I = i + (x & 1) * adjL;
int J = j + (y & 1) * adjM;
Bs[x][y][i][j] = (I < l && J < m) ? b[I][J] : 0;
}
}
}
}
lld*** s = new lld**[10];
for (int i = 0; i < 10; i++) {
switch (i) {
case 0:
s[i] = new lld*[adjL];
for (int j = 0; j < adjL; j++) {
s[i][j] = new lld[adjM];
for (int k = 0; k < adjM; k++) {
s[i][j][k] = Bs[0][1][j][k] - Bs[1][1][j][k];
}
}
break;
case 1:
s[i] = new lld*[adjN];
for (int j = 0; j < adjN; j++) {
s[i][j] = new lld[adjL];
for (int k = 0; k < adjL; k++) {
s[i][j][k] = As[0][0][j][k] + As[0][1][j][k];
}
}
break;
case 2:
s[i] = new lld*[adjN];
for (int j = 0; j < adjN; j++) {
s[i][j] = new lld[adjL];
for (int k = 0; k < adjL; k++) {
s[i][j][k] = As[1][0][j][k] + As[1][1][j][k];
}
}
break;
case 3:
s[i] = new lld*[adjL];
for (int j = 0; j < adjL; j++) {
s[i][j] = new lld[adjM];
for (int k = 0; k < adjM; k++) {
s[i][j][k] = Bs[1][0][j][k] - Bs[0][0][j][k];
}
}
break;
case 4:
s[i] = new lld*[adjN];
for (int j = 0; j < adjN; j++) {
s[i][j] = new lld[adjL];
for (int k = 0; k < adjL; k++) {
s[i][j][k] = As[0][0][j][k] + As[1][1][j][k];
}
}
break;
case 5:
s[i] = new lld*[adjL];
for (int j = 0; j < adjL; j++) {
s[i][j] = new lld[adjM];
for (int k = 0; k < adjM; k++) {
s[i][j][k] = Bs[0][0][j][k] + Bs[1][1][j][k];
}
}
break;
case 6:
s[i] = new lld*[adjN];
for (int j = 0; j < adjN; j++) {
s[i][j] = new lld[adjL];
for (int k = 0; k < adjL; k++) {
s[i][j][k] = As[0][1][j][k] - As[1][1][j][k];
}
}
break;
case 7:
s[i] = new lld*[adjL];
for (int j = 0; j < adjL; j++) {
s[i][j] = new lld[adjM];
for (int k = 0; k < adjM; k++) {
s[i][j][k] = Bs[1][0][j][k] + Bs[1][1][j][k];
}
}
break;
case 8:
s[i] = new lld*[adjN];
for (int j = 0; j < adjN; j++) {
s[i][j] = new lld[adjL];
for (int k = 0; k < adjL; k++) {
s[i][j][k] = As[0][0][j][k] - As[1][0][j][k];
}
}
break;
case 9:
s[i] = new lld*[adjL];
for (int j = 0; j < adjL; j++) {
s[i][j] = new lld[adjM];
for (int k = 0; k < adjM; k++) {
s[i][j][k] = Bs[0][0][j][k] + Bs[0][1][j][k];
}
}
break;
}
}
lld*** p = new lld**[7];
p[0] = Strassen(As[0][0], s[0], adjN, adjL, adjM);
p[1] = Strassen(s[1], Bs[1][1], adjN, adjL, adjM);
p[2] = Strassen(s[2], Bs[0][0], adjN, adjL, adjM);
p[3] = Strassen(As[1][1], s[3], adjN, adjL, adjM);
p[4] = Strassen(s[4], s[5], adjN, adjL, adjM);
p[5] = Strassen(s[6], s[7], adjN, adjL, adjM);
p[6] = Strassen(s[8], s[9], adjN, adjL, adjM);
for (int i = 0; i < adjN; i++) {
for (int j = 0; j < adjM; j++) {
c[i][j] = p[4][i][j] + p[3][i][j] - p[1][i][j] + p[5][i][j];
if (j + adjM < m)
c[i][j + adjM] = p[0][i][j] + p[1][i][j];
if (i + adjN < n)
c[i + adjN][j] = p[2][i][j] + p[3][i][j];
if (i + adjN < n && j + adjM < m)
c[i + adjN][j + adjM] = p[4][i][j] + p[0][i][j] - p[2][i][j] - p[6][i][j];
}
}
for (int x = 0; x < 2; x++) {
for (int y = 0; y < 2; y++) {
for (int i = 0; i < adjN; i++) {
delete[] As[x][y][i];
}
delete[] As[x][y];
}
delete[] As[x];
}
delete[] As;
for (int x = 0; x < 2; x++) {
for (int y = 0; y < 2; y++) {
for (int i = 0; i < adjL; i++) {
delete[] Bs[x][y][i];
}
delete[] Bs[x][y];
}
delete[] Bs[x];
}
delete[] Bs;
for (int i = 0; i < 10; i++) {
switch (i) {
case 0:
case 3:
case 5:
case 7:
case 9:
for (int j = 0; j < adjL; j++) {
delete[] s[i][j];
}
break;
case 1:
case 2:
case 4:
case 6:
case 8:
for (int j = 0; j < adjN; j++) {
delete[] s[i][j];
}
break;
}
delete[] s[i];
}
delete[] s;
for (int i = 0; i < 7; i++) {
for (int j = 0; j < (n >> 1); j++) {
delete[] p[i][j];
}
delete[] p[i];
}
delete[] p;
return c;
}
int main()
{
lld** matA;
matA = new lld*[2];
for (int i = 0; i < 2; i++)
matA[i] = new lld[3];
matA[0][0] = 1;
matA[0][1] = 2;
matA[0][2] = 3;
matA[1][0] = 4;
matA[1][1] = 5;
matA[1][2] = 6;
lld** matB;
matB = new lld*[3];
for (int i = 0; i < 3; i++)
matB[i] = new lld[2];
matB[0][0] = 7;
matB[0][1] = 8;
matB[1][0] = 9;
matB[1][1] = 10;
matB[2][0] = 11;
matB[2][1] = 12;
lld** matC = Strassen(matA, matB, 2, 3, 2);
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
printf("%lld ", matC[i][j]);
}
printf("\n");
}
return 0;
}
Output: 58 64
139 154
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