简单的dp@poj（2）2479最大子段和

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POJ 2479 Maximum sum

Description

Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below

Your task is to calculate d(A).

(1)状态：left[i]以第i位结尾时最大子串和 Right[i]为以i为间隔位之后的字符串任何组合的最大字段和

(2)方程：if(left[i-1]<0) left[i]=a[i] else left[i]=left[i-1]+a[i] 同理Right[i]

(3)初值：left[i]=right[i]=a[i]

Attention:在poj如果用cin cout 此题 TLE

#include <iostream>
#include <stdio.h>
using namespace std;
//cin cout »»³Éprintf scanf ¹ýÁË£¡
int main()
{
   int a[50001];
   int left[50001],right[50001],Right[50001];
   int Num,Length;
   scanf("%d",&Num);
   while(Num--)
   {
   scanf("%d",&Length);

   for(int i=1; i<=Length; i++)
   {
   scanf("%d",&a[i]);

   left[i]=right[i]=Right[i]=a[i];
   }
   left[0]=-1;
   right[Length+1]=-1;
   for(int i=1; i<=Length; i++)
   {
   if(left[i-1]>0)
   left[i]=left[i-1]+left[i];
   }
   int r_max=-30000;
   for(int i=Length; i>=1; i--)
   {

   if(right[i+1]>0)
   right[i]=right[i+1]+right[i];
   r_max=right[i]>r_max?right[i]:r_max;
   Right[i]=r_max;
   }
   int max=-30000;
   for(int i=1;i<Length;i++)
   {
   max=left[i]+Right[i+1]>max?left[i]+Right[i+1]:max;

   }
   printf("%d\n",max);
   }
   return 0;
}