poj----Maximum sum(poj 2479)

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30704		Accepted: 9408

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.

Source

POJ Contest,Author:Mathematica@ZSU

题意：

给一个数列，求出数列中不相交的两个字段和，要求和最大。

思路:

对每一个i来说，求出【0~i-1】的最大子段和以及【i~n-1】的最大子段和，再加起来求最大的一个就行了。[0~i-1]的最大子段和从左向右扫描，【i~n-1】从右想左扫描

即可，时间复杂度O(n).

代码：

 1 #include<iostream>
 2 #include<cstdio>
 3 #define maxn 50001
 4 #include<algorithm>
 5 using namespace std;
 6 int  a[maxn];
 7 int left[maxn];
 8 int right[maxn];
 9 int max(int a,int b)
10 {
11     return a>b?a:b;
12 }
13 int main()
14 {
15     int t,i;
16     scanf("%d",&t);
17     while(t--)
18     {
19         int n;
20       scanf("%d",&n);
21       for(i=0;i<n;i++)
22           scanf("%d",&a[i]);
23       //此时::left【i】为包涵i最大字段和
24          ::left[0]=a[0];
25       for( i=1; i<n;i++)
26           if(::left[i-1]<0)
27               ::left[i]=a[i];
28           else
29               ::left[i]=::left[i-1]+a[i];
30      //此时left[i]为i左边最大字段和
31         for(i=1; i<n;i++)
32             ::left[i]=max(::left[i],::left[i-1]);
33             ::right[n-1]=a[n-1];
34         for(i=n-2;i>=0;i--)
35         {
36             if(::right[i+1]<0)
37                 ::right[i]=a[i];
38             else
39                 ::right[i]=::right[i+1]+a[i];
40         }
41         for(i=n-2;i>=0;i--)
42             ::right[i]=max(::right[i+1],::right[i]);
43         int res=-100000000;
44         for(i=1;i<n;i++)
45         {
46             res=max(res,::left[i-1]+::right[i]);
47         }
48         printf("%dn",res);
49     }
50   return 0;
51 }