顺序表应用7：最大子段和之分治递归法

Problem Description

 给定n(1<=n<=50000)个整数（可能为负数）组成的序列a[1],a[2],a[3],…,a[n],求该序列如a[i]+a[i+1]+…+a[j]的子段和的最大值。当所给的整数均为负数时定义子段和为0，依此定义，所求的最优值为： Max{0,a[i]+a[i+1]+…+a[j]},1<=i<=j<=n。 例如，当（a[1],a[2],a[3],a[4],a[5],a[6]）=(-2,11,-4,13,-5,-2)时，最大子段和为20。

 

注意：本题目要求用分治递归法求解，除了需要输出最大子段和的值之外，还需要输出求得该结果所需的递归调用总次数。

 

递归调用总次数的获得，可以参考以下求菲波那切数列的代码段中全局变量count的用法：

#include
int count=0;
int main()
{
    int n,m;
    int fib(int n);
    scanf("%d",&n);
    m=fib(n);
    printf("%d %d\n",m,count);
    return 0;
}
int fib(int n)
{
    int s;
    count++;
    if((n==1)||(n==0)) return 1;
    else s=fib(n-1)+fib(n-2);
    return s;
}
 

Input

第一行输入整数n(1<=n<=50000)，表示整数序列中的数据元素个数；

第二行依次输入n个整数，对应顺序表中存放的每个数据元素值。

Output

一行输出两个整数，之间以空格间隔输出：

第一个整数为所求的最大子段和；

第二个整数为用分治递归法求解最大子段和时，递归函数被调用的总次数。

Example Input

6
-2 11 -4 13 -5 -2

Example Output

20 11

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define INITLISTSIZE 50000
#define INCREMENT 10
#define OVERFLOW -2
#define ERROR -1
#define OK 1

typedef int Elemtype;
typedef int Statu;
int count;
typedef struct node
{
    Elemtype *elem;
    int length;
    int listsize;
} Sqlist;

Statu Initlist(Sqlist &L);
Statu Insertlist(Sqlist &L, int i, int e);
Statu Dellist(Sqlist &L, int i);
int  Changelist(Sqlist L);
int  Divide(Sqlist L, int l, int r);
Statu Destroylist(Sqlist &L);

int main()
{
    int n, i, j;
        Sqlist L;
        Initlist(L);
        scanf("%d", &j);
        for(i = 0; i < j; i++)
        {
            int k;
            scanf("%d", &k);
            Insertlist(L, i + 1, k);
        }
        count = 0;
        n = Divide(L, 0, j - 1);
        printf("%d %d\n", n, count);
        Destroylist(L);
}

Statu Initlist(Sqlist &L)
{
    L.elem = (Elemtype *)malloc(INITLISTSIZE * sizeof(Elemtype));
    if(!L.elem)
        return OVERFLOW;
    L.length = 0;
    L.listsize = INITLISTSIZE;
    return OK;
}
Statu Insertlist(Sqlist &L, int i, int e)
{
    Elemtype *newbase;
    if(i < 1 || i > L.length + 1)
        return ERROR;
    if(L.length >= L.listsize)
    {
        newbase = (Elemtype *)realloc(L.elem, (L.listsize + INCREMENT) * sizeof(Elemtype));
        if(!newbase)
            return OVERFLOW;
        L.elem = newbase;
        L.listsize += INCREMENT;
    }
    for(int j = L.length; j >= i; j--)
        L.elem[j] = L.elem[j - 1];
    L.elem[i - 1] = e;
    ++L.length;
    return OK;
}
Statu Dellist(Sqlist &L, int i)
{
    if(i < 1 || i > L.length)
        return OVERFLOW;
    for(int j = i - 1; j < L.length; j++)
        L.elem[j] = L.elem[j + 1];
    --L.length;
    return OK;
}
int  Changelist(Sqlist L)
{
    int ThisSum, MaxSum, j;
    ThisSum = MaxSum = 0;
    for(j = 0; j < L.length; j++)
    {
       ThisSum += L.elem[j];
       if(ThisSum > MaxSum)
           MaxSum = ThisSum;
        else if(ThisSum < 0)
           ThisSum = 0;
    }
    return MaxSum;
}
Statu  Destroylist(Sqlist &L)
{
    if(L.elem)
    {
        free(L.elem);
        L.length = 0;
        L.listsize = 0;
    }
    else
        return ERROR;
    return OK;
}
int Divide(Sqlist L, int l, int r)
{
   int MaxLSum, MaxRSum;
   int MaxLBSum, MaxRBSum;
   int LBSum, RBSum;
   int Center, i;
   count++;
   if(l == r)
   {
     if(L.elem[l] > 0)
        return L.elem[l];
    else
        return 0;
   }
    Center = (l + r)/2;
    MaxLSum = Divide(L, l, Center);
    MaxRSum = Divide(L, Center + 1, r);
    MaxLBSum = 0;
    LBSum = 0;
    for(i = Center; i >= l; i--)
    {
       LBSum += L.elem[i];
       if(LBSum > MaxLBSum)
          MaxLBSum = LBSum;
    }
    MaxRBSum = 0;
    RBSum = 0;
    for(i = Center + 1; i <= r; i++)
    {
       RBSum += L.elem[i];
       if(RBSum > MaxRBSum)
          MaxRBSum = RBSum;
    }
    if(MaxLSum > MaxRSum && MaxLSum > MaxLBSum + MaxRBSum)
      return MaxLSum;
    else if(MaxRSum > MaxLSum && MaxRSum > MaxLBSum + MaxRBSum)
      return MaxRSum;
    else
      return MaxLBSum + MaxRBSum;
}