演算法設計與分析--求最大子段和問題

問題描述：

給定由n個整陣列成的序列(a1,a2, …,an)，求該序列形如

的子段和的最大值，當所有整數均為負整數時，其最大子段和為0。

利用蠻力法求解：

int maxSum(int a[],int n)
{
	int maxSum = 0;
	int sum = 0;
	for(int i = 0; i < n; i++) //從第一個數開始算起
	{
		for(int j = i + 1; j < n; j++)//從i的第二個數開始算起
		{
			sum = a[i];
			a[i]  += a[j];
			if(a[i] > sum)
			{
				sum = a[i];		//每一趟的最大值
			}
		}
		if(sum > maxSum)
		{
			maxSum = sum;
		}

	}
	return maxSum;
}
 

利用分治法求解：

int maxSum(int a[],int left, int right)
{
	int sum = 0;
	if(left == right)	//如果序列長度為1，直接求解
	{
		if(a[left] > 0) sum = a[left];
		else sum = 0;
	}
	else 
	{
		int center = (left + right) / 2;	//劃分
		int leftsum = maxSum(a,left,center);	//對應情況1，遞迴求解
		int rightsum = maxSum(a, center + 1, right);//對應情況2， 遞迴求解
		int s1 = 0;
		int lefts = 0;
		for(int i = center; i >= left; i--)	//求解s1
		{
			lefts += a[i];
			if(lefts > s1) s1 = lefts;	//左邊最大值放在s1
		}
		int s2 = 0; 
		int rights = 0;
		for(int j = center + 1; j <= right; j++)//求解s2
		{
			rights += a[j];
			if(rights > s2) s2 =rights;
		}
		sum = s1 + s2;				//計算第3鍾情況的最大子段和
		if(sum < leftsum) sum = leftsum;	//合併，在sum、leftsum、rightsum中取最大值
		if(sum < rightsum) sum = rightsum;
	}
	return sum;
}
 

利用動態規劃法求解：

int DY_Sum(int a[],int n)
{
	int sum = 0;
	int *b = (int *) malloc(n * sizeof(int));	//動態為陣列分配空間
	b[0] = a[0];
	for(int i = 1; i < n; i++)
	{
		if(b[i-1] > 0)
			b[i] = b[i - 1] + a[i];
		else
			b[i] = a[i];
	}
	for(int j = 0; j < n; j++)
	{
		if(b[j] > sum)
			sum = b[j];
	}
	delete []b;		//釋放記憶體
	return sum;
}
 

完整測試程式：

#include<iostream>
#include<time.h>
#include<Windows.h>
using namespace std;
#define MAX 10000

int BF_Sum(int a[],int n)   
{
	int max=0;     
	int sum=0;        
	int i,j;
	for (i=0;i<n-1;i++)        
	{         
		sum=a[i];          
		for(j=i+1;j<n;j++)            
		{       
			if(sum>=max)                
			{                                         
				max=sum;                
			}  
			sum+=a[j];         
		}    
	}    
	return max;
}    
int maxSum1(int a[],int left, int right)
{
	int sum = 0;
	if(left == right)	//如果序列長度為1，直接求解
	{
		if(a[left] > 0) sum = a[left];
		else sum = 0;
	}
	else 
	{
		int center = (left + right) / 2;	//劃分
		int leftsum = maxSum1(a,left,center);	//對應情況1，遞迴求解
		int rightsum = maxSum1(a, center + 1, right);//對應情況2， 遞迴求解
		int s1 = 0;
		int lefts = 0;
		for(int i = center; i >= left; i--)	//求解s1
		{
			lefts += a[i];
			if(lefts > s1) s1 = lefts;	//左邊最大值放在s1
		}
		int s2 = 0; 
		int rights = 0;
		for(int j = center + 1; j <= right; j++)//求解s2
		{
			rights += a[j];
			if(rights > s2) s2 =rights;
		}
		sum = s1 + s2;				//計算第3鍾情況的最大子段和
		if(sum < leftsum) sum = leftsum;	//合併，在sum、leftsum、rightsum中取最大值
		if(sum < rightsum) sum = rightsum;
	}
	return sum;
}

int DY_Sum(int a[],int n)
{
	int sum = 0;
	int *b = (int *) malloc(n * sizeof(int));	//動態為陣列分配空間
	b[0] = a[0];
	for(int i = 1; i < n; i++)
	{
		if(b[i-1] > 0)
			b[i] = b[i - 1] + a[i];
		else
			b[i] = a[i];
	}
	for(int j = 0; j < n; j++)
	{
		if(b[j] > sum)
			sum = b[j];
	}
	delete []b;		//釋放記憶體
	return sum;
}

int main()
{
	int num[MAX];
	int i;
	const int n = 40;
	LARGE_INTEGER begin,end,frequency;
	QueryPerformanceFrequency(&frequency);
	//生成隨機序列
	cout<<"生成隨機序列：";
	srand(time(0));
	for(int i = 0; i < n; i++)
	{
		if(rand() % 2 == 0)
			num[i] = rand();
		else
			num[i] = (-1) * rand();
		if(n < 100)
			cout<<num[i]<<" ";
	}
	cout<<endl;

	//蠻力法//
	cout<<"\n蠻力法:"<<endl;
	cout<"最大欄位和：";
	QueryPerformanceCounter(&begin);
	cout<<BF_Sum(num,n)<<endl;
	QueryPerformanceCounter(&end);
	cout<<"時間:"
		<<(double)(end.QuadPart - begin.QuadPart) / frequency.QuadPart
		<<"s"<<endl;

	cout<<"\n分治法:"<<endl;
	cout<"最大欄位和：";
	QueryPerformanceCounter(&begin);
	cout<<maxSum1(num,0,n)<<endl;
	QueryPerformanceCounter(&end);
	cout<<"時間:"
		<<(double)(end.QuadPart - begin.QuadPart) / frequency.QuadPart
		<<"s"<<endl;

	cout<<"\n動態規劃法:"<<endl;
	cout<"最大欄位和：";
	QueryPerformanceCounter(&begin);
	cout<<DY_Sum(num,n)<<endl;
	QueryPerformanceCounter(&end);
	cout<<"時間:"
		<<(double)(end.QuadPart - begin.QuadPart) / frequency.QuadPart
		<<"s"<<endl;

	system("pause");
	return 0;
}