演算法設計與分析–求最大子段和問題（蠻力法、分治法、動態規劃法) C++實現

演算法設計與分析--求最大子段和問題

問題描述：

 

給定由n個整數組成的序列(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鐘情況的最大子段和if(sum < leftsum) sum = leftsum;//合并，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));//動態為數組分配空間b[0] = a[0];for(int i = 1; i < n; i++){if(b[i-1] > 0)b[i] = b[i - 1] + a[i];elseb[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 10000int 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鐘情況的最大子段和if(sum < leftsum) sum = leftsum;//合并，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));//動態為數組分配空間b[0] = a[0];for(int i = 1; i < n; i++){if(b[i-1] > 0)b[i] = b[i - 1] + a[i];elseb[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);//產生隨機序列cout<<"產生隨機序列：";srand(time(0));for(int i = 0; i < n; i++){if(rand() % 2 == 0)num[i] = rand();elsenum[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;}

測試結果：