分治策略

最大子数组

#include<iostream>#include<climits>using namespace std;int getMiddle(int a[], int low, int mid, int high);int maxSubset(int a[], int low, int high);int main(){int a[] = { 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7 };cout << "the maxsubset:" << maxSubset(a, 0, 15) << endl;system("pause");}int getMiddle(int a[], int low, int mid, int high){int left_max = INT_MIN;int right_max = INT_MIN;int sum = 0;for (int i = mid; i >= low; i--)//计算左边的最大和{sum += a[i];if (sum > left_max) left_max = sum;}sum = 0;for (int i = mid + 1; i <= high; i++)//计算右边的最大和{sum += a[i];if (sum > right_max) right_max = sum;}return right_max + left_max;}int maxSubset(int a[], int low,int high){if (low == high)return a[low];else{int mid = (low + high) / 2;int left = maxSubset(a, 0, mid);//最大和在左边，递归调用int right = maxSubset(a, mid + 1, high);//最大和在右边，递归调用int middle = getMiddle(a, low,mid, high);//最大和在中间if (middle > left&&middle > right) return middle;else if (left > middle&&left > right) return left;else return right;}}

Strassen方法矩阵乘法

对于矩阵乘法 C = A × B，通常的做法是将矩阵进行分块相乘，如下图所示：

从上图可以看出这种分块相乘总共用了8次乘法，当然对于子矩阵相乘（如A0×B0），还可以继续递归使用分块相乘。对于中小矩阵来说，很适合使用这种分块乘法，但是对于大矩阵来说，递归的次数较多，如果能减少每次分块乘法的次数，那么性能将可以得到很好的提高。

Strassen矩阵乘法就是采用了一个简单的运算技巧，将上面的8次矩阵相乘变成了7次乘法，看别小看这减少的1次乘法，因为每递归1次，性能就提高了1/8，比如对于1024*1024的矩阵，第1次先分解成7次512*512的矩阵相乘，对于512*512的矩阵，又可以继续递归分解成256*256的矩阵相乘，…，一直递归下去，假设分解到64*64的矩阵大小后就不再递归，那么所花的时间将是分块矩阵乘法的(7/8) * (7/8) * (7/8) * (7/8) = 0.586倍，提高了快接近一倍。当然这是理论上的值，因为实际上strassen乘法增加了其他运算开销，实际性能会略低一点。

由上可见，Strassen矩阵乘法是通过递归实现的，它将一般情况下二阶矩阵乘法（可扩展到n阶，但Strassen矩阵乘法要求n是2的幂）所需的8次乘法降低为7次，其C++实现代码如下：

下面就是Strassen矩阵乘法的实现方法，

    M1 = (A0 + A3) × (B0 + B3)

   M2 = (A2 + A3) × B0

    M3 = A0 × (B1 - B3)

    M4 = A3 × (B2 - B0)

    M5 = (A0 + A1) × B3

    M6 = (A2 - A0) × (B0 + B1)

    M7 = (A1 - A3) × (B2 + B3)

    C0 = M1 + M4 - M5 + M7

    C1 = M3 + M5

    C2 = M2 + M4

    C3 = M1 - M2 + M3 + M6

在求解M1,M2,M3,M4,M5,M6,M7时需要使用7次矩阵乘法，其他都是矩阵加法和减法。

下面看看Strassen矩阵乘法的串行实现伪代码：

Serial_StrassenMultiply(A, B, C)

{

    T1 = A0 + A3;

    T2 = B0 + B3;

    StrassenMultiply(T1, T2, M1);

    T1 = A2 + A3;

    StrassenMultiply(T1, B0, M2);

    T1 = (B1 - B3);

    StrassenMultiply(A0, T1, M3);

 

    T1 = B2 - B0;

    StrassenMultiply(A3, T1, M4);

 

   T1 = A0 + A1;

   StrassenMultiply(T1, B3, M5);       

   

    T1 = A2 – A0;

    T2 = B0 + B1;

    StrassenMultiply(T1, T2, M6);

    T1 = A1 – A3;

    T2 = B2 + B3;

    StrassenMultiply(T1, T2, M7);

    C0 = M1 + M4 - M5 + M7

    C1 = M3 + M5

    C2 = M2 + M4

    C3 = M1 - M2 + M3 + M6

}

#include <iostream>  using namespace std;const int N = 6;     //Define the size of the Matrix  template<typename T>void Strassen(int n, T A[][N], T B[][N], T C[][N]);template<typename T>void input(int n, T p[][N]);template<typename T>void output(int n, T C[][N]);int main() {//Define three Matrices  int A[N][N], B[N][N], C[N][N];//对A和B矩阵赋值，随便赋值都可以，测试用  for (int i = 0; i<N; i++) {for (int j = 0; j<N; j++) {A[i][j] = i * j;B[i][j] = i * j;}}//调用Strassen方法实现C=A*B  Strassen(N, A, B, C);//输出矩阵C中值  output(N, C);system("pause");return 0;}template<typename T>void input(int n, T p[][N]) {for (int i = 0; i<n; i++) {cout << "Please Input Line " << i + 1 << endl;for (int j = 0; j<n; j++) {cin >> p[i][j];}}}template<typename T>void output(int n, T C[][N]) {cout << "The Output Matrix is :" << endl;for (int i = 0; i<n; i++) {for (int j = 0; j<n; j++) {cout << C[i][j] << "" << endl;}}}template<typename T>void Matrix_Multiply(T A[][N], T B[][N], T C[][N]) {  //Calculating A*B->C  for (int i = 0; i<2; i++) {for (int j = 0; j<2; j++) {C[i][j] = 0;for (int t = 0; t<2; t++) {C[i][j] = C[i][j] + A[i][t] * B[t][j];}}}}template <typename T>void Matrix_Add(int n, T X[][N], T Y[][N], T Z[][N]) {for (int i = 0; i<n; i++) {for (int j = 0; j<n; j++) {Z[i][j] = X[i][j] + Y[i][j];}}}template <typename T>void Matrix_Sub(int n, T X[][N], T Y[][N], T Z[][N]) {for (int i = 0; i<n; i++) {for (int j = 0; j<n; j++) {Z[i][j] = X[i][j] - Y[i][j];}}}template <typename T>void Strassen(int n, T A[][N], T B[][N], T C[][N]) {T A11[N][N], A12[N][N], A21[N][N], A22[N][N];T B11[N][N], B12[N][N], B21[N][N], B22[N][N];T C11[N][N], C12[N][N], C21[N][N], C22[N][N];T M1[N][N], M2[N][N], M3[N][N], M4[N][N], M5[N][N], M6[N][N], M7[N][N];T AA[N][N], BB[N][N];if (n == 2) {  //2-order  Matrix_Multiply(A, B, C);}else {//将矩阵A和B分成阶数相同的四个子矩阵，即分治思想。  for (int i = 0; i<n / 2; i++) {for (int j = 0; j<n / 2; j++) {A11[i][j] = A[i][j];A12[i][j] = A[i][j + n / 2];A21[i][j] = A[i + n / 2][j];A22[i][j] = A[i + n / 2][j + n / 2];B11[i][j] = B[i][j];B12[i][j] = B[i][j + n / 2];B21[i][j] = B[i + n / 2][j];B22[i][j] = B[i + n / 2][j + n / 2];}}//Calculate M1 = (A0 + A3) × (B0 + B3)  Matrix_Add(n / 2, A11, A22, AA);Matrix_Add(n / 2, B11, B22, BB);Strassen(n / 2, AA, BB, M1);//Calculate M2 = (A2 + A3) × B0  Matrix_Add(n / 2, A21, A22, AA);Strassen(n / 2, AA, B11, M2);//Calculate M3 = A0 × (B1 - B3)  Matrix_Sub(n / 2, B12, B22, BB);Strassen(n / 2, A11, BB, M3);//Calculate M4 = A3 × (B2 - B0)  Matrix_Sub(n / 2, B21, B11, BB);Strassen(n / 2, A22, BB, M4);//Calculate M5 = (A0 + A1) × B3  Matrix_Add(n / 2, A11, A12, AA);Strassen(n / 2, AA, B22, M5);//Calculate M6 = (A2 - A0) × (B0 + B1)  Matrix_Sub(n / 2, A21, A11, AA);Matrix_Add(n / 2, B11, B12, BB);Strassen(n / 2, AA, BB, M6);//Calculate M7 = (A1 - A3) × (B2 + B3)  Matrix_Sub(n / 2, A12, A22, AA);Matrix_Add(n / 2, B21, B22, BB);Strassen(n / 2, AA, BB, M7);//Calculate C0 = M1 + M4 - M5 + M7  Matrix_Add(n / 2, M1, M4, AA);Matrix_Sub(n / 2, M7, M5, BB);Matrix_Add(n / 2, AA, BB, C11);//Calculate C1 = M3 + M5  Matrix_Add(n / 2, M3, M5, C12);//Calculate C2 = M2 + M4  Matrix_Add(n / 2, M2, M4, C21);//Calculate C3 = M1 - M2 + M3 + M6  Matrix_Sub(n / 2, M1, M2, AA);Matrix_Add(n / 2, M3, M6, BB);Matrix_Add(n / 2, AA, BB, C22);//Set the result to C[][N]  for (int i = 0; i<n / 2; i++) {for (int j = 0; j<n / 2; j++) {C[i][j] = C11[i][j];C[i][j + n / 2] = C12[i][j];C[i + n / 2][j] = C21[i][j];C[i + n / 2][j + n / 2] = C22[i][j];}}}}