poj-2479

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#poj #动态规划 #2479

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本文介绍了一种通过动态规划求解最大子数组和问题的方法，包括从前向后和从后向前两次遍历数组来确定任意两个子数组的最大和。文章提供了一个具体的样例及其解析过程，并附带了完整的C++实现代码。

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Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 42649		Accepted: 13275

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).

Input

The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.

Output

Print exactly one line for each test case. The line should contain the integer d(A).

Sample Input

1

10
1 -1 2 2 3 -3 4 -4 5 -5

Sample Output

Hint

In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.

Huge input,scanf is recommended.

/*动态规划
从前向后遍历一次，找到以i结束的最大和
从后向前遍历一次，找到以i开始的最大和
遍历一遍找到两段的最大和*/
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
#include<stdio.h>
#define MAX 200000
#define max(A,B)((A)>=(B)?(A):(B))
const int inf=-999999;
int l[MAX],r[MAX],a[MAX],dp1[MAX],dp2[MAX];
int main()
{
 int i,t,n,maxsum;
    scanf("%d\n",&t);//案例组
    while(t--){
        scanf("%d",&n);//每组案例数
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);//把案例的数都存放到数组里
  dp1[1]=a[1];
  dp2[n]=a[n];
  l[1]=a[1];
  r[n]=a[n];
  for(i=2;i<=n;i++)//从左遍历，找最大和
  {
      dp1[i]=max(dp1[i-1]+a[i],a[i]);
      l[i]=max(dp1[i],l[i-1]);
      }
  for(i=n-1;i>=1;i--)//从右遍历，找最大和
  {
      dp2[i]=max(dp2[i+1]+a[i],a[i]);
      r[i]=max(dp2[i],r[i+1]);
      }
  maxsum=inf;
  for(i=1;i<=n-1;i++){//找这两段的最大和
      maxsum=max(maxsum,l[i]+r[i+1]);
  }
  printf("%d\n",maxsum);
}
 return 0;
}