算法导论--第四章:分治策略

4.1 最大子数组问题

一个数组arr长度为len,假定0<=i<=j<len,求arr[ j ] - arr[ i ]的最大值.

暴力求解方法,复杂度为n*n:

#include <stdio.h>

int max_sub( int arr[], int len )
{
	int i = 0;
	int j = 0;
	int maxValue = 0;
	for ( i = 0; i < len - 1; i++ ){
		for ( j = i + 1; j < len; j++ ){
			if ( arr[ j ] - arr[ i ] > maxValue ){
				maxValue = arr[ j ] - arr[ i ];
			}
		}
	}

	return maxValue;
}

int main( void )
{
	int arr[ 100 ];
	int i = 0;
	int result = 0;
	for ( i = 0; i < 50; i++ ){
		arr[ i ] = -i;
	}
	for ( i = 0; i < 50; i++ ){
		arr[ i + 50 ] = i;
	}

	result = max_sub( arr, 100 );

	printf("%d\n", result );

	return 0;
}

程序输出:

分治法:

#include <stdio.h>
#include <limits.h>

int find_max_crossing_subarray( int arr[], int low, int mid, int high, int *left_or_right_low, int *left_or_right_high )
{
	int left_sum = INT_MIN;
	int right_sum = INT_MIN;
	int sum = 0;
	int i = 0;
	int max_left;
	int max_right;
	for ( i = mid; i >= low; i-- ){
		sum += arr[ i ];
		if ( sum > left_sum ){
			left_sum = sum;
			max_left = i;
		}
	}
	sum = 0;
	for ( i = mid + 1; i <= high; i++ ){
		sum += arr[ i ];
		if ( sum > right_sum ){
			right_sum = sum;
			max_right = i;
		}
	}
	*left_or_right_low = max_left;
	*left_or_right_high = max_right;

	return left_sum + right_sum;
}

int find_maximum_subarray( int arr[], int low, int high, int *left_or_right_low, int *left_or_right_high )
{
	int mid;
	int left_sum;
	int right_sum;
	int cross_sum;
	int left_low, left_high, right_low, right_high, cross_low, cross_high;
	if ( low == high ){
		return arr[ low ];
	}
	else{
		mid = ( low + high ) / 2;
		left_sum = find_maximum_subarray( arr, low, mid, &left_low, &left_high );
		right_sum = find_maximum_subarray( arr, mid + 1, high, &right_low, &right_high );
		cross_sum = find_max_crossing_subarray( arr, low, mid, high, &cross_low, &cross_high );
		if ( left_sum >= right_sum && left_sum >= cross_sum ){
			*left_or_right_low = left_low;
			*left_or_right_high = left_high;
			return left_sum;
		}
		else if ( right_sum >= left_sum && right_sum >= cross_sum ){
			*left_or_right_low = right_low;
			*left_or_right_high = right_high;
			return right_sum;
		}
		else{
			*left_or_right_low = cross_low;
			*left_or_right_high = cross_high;
			return cross_sum;
		}
	}
}

int main( void )
{
	int arr[ 16 ] = { 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7 };
	int left;
	int right;
	int sum = 0;
	sum = find_maximum_subarray( arr, 0, 15, &left, &right );

	printf("%d--%d : %d\n", left, right, sum );

	return 0;
}

程序输出:

具体分析请参考书本上的讲解.

习题4.1-3:

对于此习题,由于分治法处理的数组和暴力求解法处理的数组完全不一样(虽然分治法的数组出自暴力求解法),所以将两个算法合并起来,实际上增加了难度和算法的复杂度.....

习题4.1-4:

可以用INT_MIN来初始化数组,返回为0.若返回为0,则要么为空数组,要么就是我们要找的答案,没任何冲突.

习题4.1-5:

下列代码参考以下网址:

http://www.cnblogs.com/Jiajun/archive/2013/05/08/3066979.html

#include <stdio.h>
#include <limits.h>

int find_maximum_subarray( int arr[], int low, int high, int *left, int *last )
{
	int i = 0;
	int maxValue = INT_MIN;
	int sum = 0;
	int cur_low = 0;
	for ( i = low; i < high; i++ ){
		sum += arr[ i ];
		if ( sum > maxValue ){
			*left = cur_low;
			maxValue = sum;
			*last = i;
		}
		else if ( sum < 0 ){		//当sum < 0的时候,说明最大子数组不包括之前的子数组,直接抛弃
			sum = 0;
			cur_low = i + 1;
		}
	}
	return maxValue;
}
int main( void )
{
	int arr[ 16 ] = { 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7 };
	int temparr[ 16 ];
	int sum = 0;
	int left;
	int last;
	int i = 0;
	for ( i = 0; i < 16; i++ ){
		temparr[ i ] = -i - 1;
	}
	sum = find_maximum_subarray( arr, 0, 16, &left, &last );

	printf("%d--%d: %d\n", left, last, sum );

	return 0;
}

程序输出:

4.2 矩阵乘法的Strassen算法

我们一般的矩阵相乘都类似于下列的代码:

#include <stdio.h>

int main( void )
{
	int arr1[ 2 ][ 2 ] = { 2, 3, 4, 5 };
	int arr2[ 2 ][ 2 ] = { 2, 3, 4, 5 };
	int arr3[ 2 ][ 2 ];
	int i = 0;
	int j = 0;
	int k = 0;

	for ( i = 0; i < 2; i++ ){
		for ( j = 0; j < 2; j++ ){
			arr3[ i ][ j ] = 0;
			for ( k = 0; k < 2; k++ ){
				arr3[ i ][ j ] += arr1[ i ][ k ] * arr2[ k ][ j ];
			}
		}
	}

	for ( i = 0; i < 2; i++ ){
		for ( j = 0; j < 2; j++ ){
			printf("%d ", arr3[ i ][ j ] );
		}
		printf("\n");
	}

	return 0;
}

程序输出:

而strassen的算法代码如下:

#include <stdio.h>

int main( void )
{
	int i = 0;
	int j = 0;
	int A[ 2 ][ 2 ] = { 2, 3, 4, 5 };
	int B[ 2 ][ 2 ] = { 2, 3, 4, 5 };
	int C[ 2 ][ 2 ];
	int s1 = B[ 0 ][ 1 ] - B[ 1 ][ 1 ];
	int s2 = A[ 0 ][ 0 ] + A[ 0 ][ 1 ];
	int s3 = A[ 1 ][ 0 ] + A[ 1 ][ 1 ];
	int s4 = B[ 1 ][ 0 ] - B[ 0 ][ 0 ];
	int s5 = A[ 0 ][ 0 ] + A[ 1 ][ 1 ];
	int s6 = B[ 0 ][ 0 ] + B[ 1 ][ 1 ];
	int s7 = A[ 0 ][ 1 ] - A[ 1 ][ 1 ];
	int s8 = B[ 1 ][ 0 ] + B[ 1 ][ 1 ];
	int s9 = A[ 0 ][ 0 ] - A[ 1 ][ 0 ];
	int s10 = B[ 0 ][ 0 ] + B[ 0 ][ 1 ];
	int p1 = A[ 0 ][ 0 ] * s1;
	int p2 = s2 * B[ 1 ][ 1 ];
	int p3 = s3 * B[ 0 ][ 0 ];
	int p4 = A[ 1 ][ 1 ] * s4;
	int p5 = s5 * s6;
	int p6 = s7 * s8;
	int p7 = s9 * s10;

	C[ 0 ][ 0 ] = p5 + p4 - p2 + p6;
	C[ 0 ][ 1 ] = p1 + p2;
	C[ 1 ][ 0 ] = p3 + p4;
	C[ 1 ][ 1 ] = p5 + p1 - p3 - p7;

	for ( i = 0; i < 2; i++ ){
		for ( j = 0; j < 2; j++ ){
			printf("%d ", C[ i ][ j ] );
		}
		printf("\n");
	}

	return 0;
}

程序输出:

但是如果n不是等于2,则不知道如何进行编写代码..........

习题4.2-7:

( a + bi ) * ( c + di ) = ac - bd + ( ad + bc )i,但是bc = ac * bc / ad算出,所以只要3步即可.