算法设计与分析–求最大子段和问题
问题描述：
给定由n个整数组成的序列(a1,a2, …,an)，，求该序列形如
的子段和的最大值，当所有整数均为负整数时，其最大子段和为0。
利用蛮力法求解：
int maxSum(int a[],int n){int maxSum = 0;int sum = 0;for(int i = 0; i < n; i++) //从第一个数开始算起{for(int j = i + 1; j < n; j++)//从i的第二个数开始算起{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)//如果序列长度为1，直接求解{if(a[left] > 0) sum = a[left];else sum = 0;}else {int center = (left + right) / 2;//划分int leftsum = maxSum(a,left,center);//对应情况1，递归求解int rightsum = maxSum(a, center + 1, right);//对应情况2， 递归求解int s1 = 0;int lefts = 0;for(int i = center; i >= left; i–)//求解s1{lefts += a[i];if(lefts > s1) s1 = lefts;//左边最大值放在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;//计算第3钟情况的最大子段和if(sum < leftsum) sum = leftsum;//合并，在sum、leftsum、rightsum中取最大值if(sum < rightsum) sum = rightsum;}return sum;}
利用动态规划法求解：
int DY_Sum(int a[],int n){int sum = 0;int *b = (int *) malloc(n * sizeof(int));//动态为数组分配空间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;//释放内存return sum;}
完整测试程序：
#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)//如果序列长度为1，直接求解{if(a[left] > 0) sum = a[left];else sum = 0;}else {int center = (left + right) / 2;//划分int leftsum = maxSum1(a,left,center);//对应情况1，递归求解int rightsum = maxSum1(a, center + 1, right);//对应情况2， 递归求解int s1 = 0;int lefts = 0;for(int i = center; i >= left; i–)//求解s1{lefts += a[i];if(lefts > s1) s1 = lefts;//左边最大值放在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;//计算第3钟情况的最大子段和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));//动态为数组分配空间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;//释放内存return sum;}int main(){int num[MAX];int i;const int n = 40;LARGE_INTEGER begin,end,frequency;QueryPerformanceFrequency(&frequency);//生成随机序列cout<<"生成随机序列：";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;//蛮力法//cout<<"\n蛮力法:"<<endl;cout<"最大字段和：";QueryPerformanceCounter(&begin);cout<<BF_Sum(num,n)<<endl;QueryPerformanceCounter(&end);cout<<"时间:"<<(double)(end.QuadPart – begin.QuadPart) / frequency.QuadPart<<"s"<<endl;cout<<"\n分治法:"<<endl;cout<"最大字段和：";QueryPerformanceCounter(&begin);cout<<maxSum1(num,0,n)<<endl;QueryPerformanceCounter(&end);cout<<"时间:"<<(double)(end.QuadPart – begin.QuadPart) / frequency.QuadPart<<"s"<<endl;cout<<"\n动态规划法:"<<endl;cout<"最大字段和：";QueryPerformanceCounter(&begin);cout<<DY_Sum(num,n)<<endl;QueryPerformanceCounter(&end);cout<<"时间:"<<(double)(end.QuadPart – begin.QuadPart) / frequency.QuadPart<<"s"<<endl;system("pause");return 0;}测试结果：
灿烂甜美!那一瞬的激-情绽放,催人奋进!胜利,永远属于为梦想奋斗的人新乐吧
