hdu1006（最大连续子序列）

最新推荐文章于 2020-09-30 16:10:46 发布

原创最新推荐文章于 2020-09-30 16:10:46 发布 · 213 阅读

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本文详细解析了最大连续子序列和问题的算法实现，通过动态规划的方法找到给定序列中具有最大和的连续子序列，并输出该子序列的起始和结束位置。文章包含完整的AC代码示例，适用于解决类似HDU1231的问题。

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Given a sequence a[1],a[2],a[3]…a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Sample Output
Case 1:
14 1 4

Case 2:
7 1 6

与hdu1231的题思路一样，只是变成了输出最大连续子序列的开始序数和最后序数

最大连续子序列和公式：设dp[i]为以第i个元素结尾的连续和的最大值
则有
dp[0]=a[0]
dp[i]=max(dp[i-1]+a[i],a[i])
所以最大连续子序列和就是所有dp中最大的一个
ac代码

#include <iostream>
	using namespace std;
	int a[100005], p[100005];
	int main()
	{
		int f[20][3];
		int k,b, e,h=0;
		cin >> e;
		while (e)
		{
			cin >> k;
		
			 b = 0;
			for (int i = 0; i < k; i++)
			{
				cin >> a[i];
			}
			
			
			p[0] = a[0]; f[h][0] = a[0]; f[h][1] = 0; f[h][2] = 0;
				for (int i = 1; i < k; i++)
				{

					if (p[i - 1] + a[i] >=a[i])
						p[i] = p[i - 1] + a[i];
					else { p[i] = a[i]; b= i; }
					if (p[i] > f[h][0])
					{
						f[h][0] = p[i]; f[h][1] = i; f[h][2] = b;
					}
				}
	
				
				  h++;
				  e--;
		}
		for (int j = 1; j <= h; j++)
		{
			cout << "Case " << j << ":" << endl;
			cout << f[j - 1][0] << " " << f[j - 1][2] + 1 << " " << f[j - 1][1] + 1;
			if (j != h)cout << endl<<endl;
			else cout << endl;
		}
	}