1007 Maximum Subsequence Sum

本文探讨了最大子列求和问题的解决方案,通过一个具体示例详细讲解了如何找到具有最大元素总和的连续子序列。文章提供了完整的代码实现,并强调了处理所有数均为负数的特殊情况。

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Given a sequence of K integers N ​ 1 ​ ​ , N ​ 2 ​ ​ , . . . , N K ​ ​ { N​_1​​, N​_2​​, ..., N_K​​ } N1,N2,...,NK. A continuous subsequence is defined to be N i ​ ​ , N ​ i + 1 ​ ​ , . . . , N ​ j ​ ​ { N_i​​, N​_{i+1​}​, ..., N​_j​​ } Ni,Ni+1,...,Nj where 1 ≤ i ≤ j ≤ K 1≤i≤j≤K 1ijK. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence − 2 , 11 , − 4 , 13 , − 5 , − 2 { -2, 11, -4, 13, -5, -2 } 2,11,4,13,5,2, its maximum subsequence is 11 , − 4 , 13 { 11, -4, 13 } 11,4,13 with the largest sum being 20 20 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K ( ≤ 10000 ) K (≤10000) K(10000). The second line contains K K K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i i i and j j j (as shown by the sample case). If all the K K K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4


题解:

最大子列求和问题,只需进行简单改编即可,注意题目中对于最大和为负情况的处理;
附:子序列问题参考


#include <iostream>
using namespace std;

int main()
{
	int k;
	cin >> k;
	int *arr = new int[k];
	int ThisSum = 0, MaxSum = -1;   // 最小值为 0,故初始值应为负
	int first, last;
	int c = 0;

	for (int i = 0; i < k; i++)
		cin >> arr[i];

	for (int i = 0; i < k; i++)
	{
		ThisSum += arr[i];
		c++;
		if (ThisSum > MaxSum)
		{
			MaxSum = ThisSum;
			first = arr[i - c + 1];
			last = arr[i];
		}

		else if (ThisSum < 0)
		{
			ThisSum = 0;
			c = 0;
		}
	}
	if (MaxSum == -1)
		cout << "0" << " " << arr[0] << " " << arr[k-1] << endl;
	else
		cout << MaxSum << " " << first << " " << last << endl;
}
内容概要:本文档详细介绍了基于MATLAB实现的无人机三维路径规划项目,核心算法采用蒙特卡罗树搜索(MCTS)。项目旨在解决无人机在复杂三维环境中自主路径规划的问题,通过MCTS的随机模拟与渐进式搜索机制,实现高效、智能化的路径规划。项目不仅考虑静态环境建模,还集成了障碍物检测与避障机制,确保无人机飞行的安全性和效率。文档涵盖了从环境准备、数据处理、算法设计与实现、模型训练与预测、性能评估到GUI界面设计的完整流程,并提供了详细的代码示例。此外,项目采用模块化设计,支持多无人机协同路径规划、动态环境实时路径重规划等未来改进方向。 适合人群:具备一定编程基础,特别是熟悉MATLAB和无人机技术的研发人员;从事无人机路径规划、智能导航系统开发的工程师;对MCTS算法感兴趣的算法研究人员。 使用场景及目标:①理解MCTS算法在三维路径规划中的应用;②掌握基于MATLAB的无人机路径规划项目开发全流程;③学习如何通过MCTS算法优化无人机在复杂环境中的飞行路径,提高飞行安全性和效率;④为后续多无人机协同规划、动态环境实时调整等高级应用打下基础。 其他说明:项目不仅提供了详细的理论解释和技术实现,还特别关注了实际应用中的挑战和解决方案。例如,通过多阶段优化与迭代增强机制提升路径质量,结合环境建模与障碍物感知保障路径安全,利用GPU加速推理提升计算效率等。此外,项目还强调了代码模块化与调试便利性,便于后续功能扩展和性能优化。项目未来改进方向包括引入深度强化学习辅助路径规划、扩展至多无人机协同路径规划、增强动态环境实时路径重规划能力等,展示了广阔的应用前景和发展潜力。
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