算法导论(c++实现）chapter4

算法导论的第四章——分治策略，主要的算法为寻找最大子数组问题，还有矩阵乘法的Strassen算法，还有一些课后的练习题，主要讲了求解递归式的三种方法：代入法（感觉像是猜测，然后用数学归纳法验证），递归树法（比较的直观明了），主方法（给出了递归式的大部分通解）。现在把该chapter的所有涉及的算法都贴上来。

#include<iostream>
using namespace std;//this is the program that in order to find the max crossing subarray.
//in order to return the subarray's low bound and the high bound,
//i have to create a struct to contain all this information
//this is the constrction of the struct
template<class T>
struct SUB_Array {int low;int high;T sum;
};//this is the realization of FIND-MAX-CROSSING-SUBARRAY
template<class T>
SUB_Array<T> F_M_C_S(T* A, int low, int mid, int high)
{T left_sum = -99999;T sum = 0;int max_left;for (int i = mid; i >= low; i--){sum += A[i];if (sum > left_sum){left_sum = sum;max_left = i;}}T right_sum = -9999999;int max_right;sum = 0;for (int j = mid + 1; j <= high; j++){sum += A[j];if (sum > right_sum){right_sum = sum;max_right = j;}}SUB_Array<T> x;x.low = max_left;x.high = max_right;x.sum = left_sum + right_sum;return  x;
}//this is the FIND-MAXIMUM-SUBBARRAY
template<class T>
SUB_Array<T> F_M_S(T* A, int low, int high)
{SUB_Array<T> x,y,z;if(high==low){x.low = low;x.high = high;x.sum = A[low];return x;}else{int   mid = (low + high) / 2;x = F_M_S(A, low, mid);y = F_M_S(A, mid + 1, high);z = F_M_C_S(A, low, mid, high);if (x.sum >= y.sum&&x.sum >= z.sum)return x;else if (y.sum >= x.sum&&y.sum >= z.sum)return y;elsereturn z;}
}//solution to 4.1-2 . so i will use the violent way to solve the max subarray;template<class T>SUB_Array<T> Viol_Maxarray(T* A, int low, int high){SUB_Array<T> x;T Max_sum = A[low];int Max_low, Max_high;for (int i = low; i <= high; i++){T sum = 0;for (int j = i; j <= high; j++){sum += A[j];if (sum > Max_sum){Max_sum = sum;Max_low = i;Max_high = j;}}}x.sum = Max_sum;x.low = Max_low;x.high = Max_high;return x;}//solution to 4.1-4//in my opinion,i think the biggest subarray is that the sum must more than zero,and if the sum \// is less than zero,it just returns the zero ,which means not have.//  and it is easy to change the program ,just use the upper program,and make the left_sum=0 ,right_sum=0;//solution to 4.1-5template<class T>SUB_Array<T> Nrecurr_SUBARR(T* A, int low, int high){SUB_Array<T> x;int t_high;T Max_sum = A[low];int t_low=low;for (int i = low; i <= high; i++){T sum = 0;for (int j = i + 1; j >= t_low; j--){sum += A[j];if (sum > Max_sum){Max_sum = sum;t_low = j;t_high = i + 1;}}}x.low = t_low;x.high = t_high;x.sum = Max_sum;return x;}//this is the SQUARE-MATRIX-MULTIPLYconst int n = 4;template<class T>void SQMA_MUL(T a[n][n], T b[n][n], T c[n][n]){for(int i=0;i<n;i++)for (int j = 0; j < n; j++){c[i][j] = 0;for (int k = 0; k < n; k++)c[i][j] += a[i][k] * b[k][j];}}//this is the Strassen meathod//we just assure the input number is N*N,so i just split into n/2 * n/2.//and i didn't have much more simple coded way to realize it.//so i just use the ordinary way .//there are several steps to make ahead,so it will be sort of simple.const int Size = 4;template<class T>void Strassen(T a[Size][Size], T b[Size][Size], T c[Size][Size]){const int newsize = Size/2;T S1[newsize][newsize] = { 0 }, S2[newsize][newsize] = { 0 }, S3[newsize][newsize] = { 0 };T S4[newsize][newsize] = { 0 }, S5[newsize][newsize] = { 0 }, S6[newsize][newsize] = { 0 }, S7[newsize][newsize] = { 0 }, S8[newsize][newsize] = { 0 };T S9[newsize][newsize] = { 0 }, S10[newsize][newsize] = { 0 };for (int i = 0; i < newsize; i++)for (int j = 0; j < newsize; j++){S1[i][j] = b[i][j+newsize] - b[newsize + i][newsize + j];S2[i][j] = a[i][j] + a[i][newsize + j];S3[i][j] = a[newsize + i][j] + a[newsize + i][newsize + j];S4[i][j] = b[newsize + i][j] - b[i][j];S5[i][j] = a[i][j] + a[newsize + i][newsize + j];S6[i][j] = b[i][j] + b[newsize + i][newsize + j];S7[i][j] = a[i][newsize + j] - a[newsize + i][newsize + j];S8[i][j] = b[newsize + i][j] + b[newsize + i][newsize + j];S9[i][j] = a[i][j] - a[newsize + i][j];S10[i][j] = b[i][j] + b[i][newsize + j];}T P1[newsize][newsize] = { 0 }, P2[newsize][newsize] = { 0 }, P3[newsize][newsize] = { 0 }, P4[newsize][newsize] = { 0 };T P5[newsize][newsize] = { 0 }, P6[newsize][newsize] = { 0 }, P7[newsize][newsize] = { 0 };for (int i = 0; i < newsize; i++)for (int j = 0; j < newsize; j++){for (int k = 0; k < newsize; k++){P1[i][j] += a[i][k] * S1[k][j];P2[i][j] += S2[i][k] * b[newsize + k][newsize + j];P3[i][j] += S3[i][k] * b[k][j];P4[i][j] += a[newsize + i][newsize + k] * S4[k][j];P5[i][j] += S5[i][k] * S6[k][j];P6[i][j] += S7[i][k] * S8[k][j];P7[i][j] += S9[i][k] * S10[k][j];}}for (int i = 0; i < newsize; i++)for (int j = 0; j < newsize; j++){c[i][j] = 0;c[i][j] = P5[i][j] + P4[i][j] - P2[i][j] + P6[i][j];c[i][newsize + j] = 0;c[i][newsize + j] = P1[i][j] + P2[i][j];c[i + newsize][newsize + j] = 0;c[i + newsize][j + newsize] = P5[i][j] + P1[i][j] - P3[i][j] - P7[i][j];c[newsize + i][j] = 0;c[newsize + i][j] = P4[i][j] + P3[i][j];}}int main(){//int A[16] = { 13,-3,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7 };//SUB_Array<int> x;//x = F_M_S(A, 0, 15);//x = Viol_Maxarray(A, 0, 15);//x = Nrecurr_SUBARR(A, 0, 15);//cout << x.low <<"   "<< x.high <<"   "<< x.sum << endl;int a[4][4] = { 1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2 };int b[4][4] = { 1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2 };int c[4][4] = { 0 };int d[4][4];int e[2][2] = { 1,3,7,5 };int f[2][2] = { 6,8,4,2 };// SQMA_MUL(e, f, c);SQMA_MUL(a, b, d);Strassen<int>(a, b, c);for (int i = 0; i < n; i++){for (int j = 0; j < n; j++)cout << c[i][j] << "  ";cout << endl;}cout << endl;for (int i = 0; i < n; i++){   for (int j = 0; j < n; j++)cout << d[i][j] << "  ";cout << endl;}return 0;
}

总的算法就那么多，但是有些东西可以继续完善。有些函数的接口有待斟酌，还有就是Strassen算法，我感觉我的实现还是O(n^3)，递归实现我暂时没有想出来。先上传了，等有思路在补上！欢迎大家提供思路，一起探讨~

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