poj2479 && poj2593Maximum sum（求两个不相交最大字段的和）

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题目链接：http://poj.org/problem?id=2479

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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

1101 -1 2 2 3 -3 4 -4 5 -5

Sample Output

13

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

代码如下：

#include <cstdio>#include <iostream>using namespace std;#define INF 0x3fffffff#define M 100000+17int a[M],b[M];int main(){int n, i, T;while(~scanf("%d",&T)){while(T--){scanf("%d",&n);int sum = 0, MAX = -INF;for(i = 1; i <= n; i++){scanf("%d",&a[i]);sum+=a[i];if(sum > MAX){MAX = sum;}b[i] = MAX;if(sum < 0){sum = 0;}}MAX = -INF;sum = 0;int ans = MAX, t;for(i = n; i > 1; i--){sum+=a[i];if(sum > MAX){MAX = sum;}t = MAX + b[i-1];if(t > ans){ans = t;}if(sum < 0){sum = 0;}}printf("%d\n",ans);}}return 0;}