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 -5Sample Output
13Hint
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("%d\n",res);
49 }
50 return 0;
51 }腾讯云开发者
