poj2479 c++ : Maximum sum

poj2479 c++ : Maximum sum

描述

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

                     t1     t2 
         d(A) = max{ ∑ai + ∑aj | 1 <= s1 <= t1 < s2 <= t2 <= n }
                    i=s1   j=s2

Your task is to calculate d(A).

輸入

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.

輸出

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

樣例輸入

1

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

樣例輸出

13

題目大意：

給出一個數組，找到兩段和最大，i<=j

解題思路：

動態規劃DP問題，從前向後，從後向前遍歷兩遍，分別找到以i爲結束和以i爲開始的最大和，最後再遍歷一遍找到兩段的最大和，

#include<stdio.h>
#include<iostream>
#define max(A,B)((A)>=(B)?(A):(B))
using namespace std;

const int INF=100005;
int a[INF];
int lt[INF];
int rt[INF];
int dp1[INF];
int dp2[INF];
int main()
{
 int i,j,c,n,sum;
 scanf("%d",&c);
 while(c--)
 {
  scanf("%d",&n);
  for(i=1;i<=n;i++)
   scanf("%d",&a[i]);
  dp1[1]=a[1];
  dp2[n]=a[n];
  lt[1]=a[1];
  rt[n]=a[n];

  for(i=2;i<=n;i++)
  {dp1[i]=max(dp1[i-1]+a[i],a[i]);
  lt[i]=max(dp1[i],lt[i-1]);}

  for(i=n-1;i>=1;i--)
  {dp2[i]=max(dp2[i+1]+a[i],a[i]);
  rt[i]=max(dp2[i],rt[i+1]);}

  sum=-INF;
  for(i=1;i<=n-1;i++){
      sum=max(sum,lt[i]+rt[i+1]);
  }
  printf("%d\n",sum);
 }
 return 0;
}