HDU-1231&HDU-1003-优快云博客

本文链接：https://blog.youkuaiyun.com/Lin_ZR/article/details/74620554

本文介绍了一种解决最大连续子序列问题的方法，通过动态规划实现了寻找给定序列中具有最大和的连续子序列及其起始和终止元素的过程。

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最大连续子序列

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32827 Accepted Submission(s): 14770

Problem Description

给定K个整数的序列{ N1, N2, ..., NK }，其任意连续子序列可表示为{ Ni, Ni+1, ...,
Nj }，其中 1 <= i <= j <= K。最大连续子序列是所有连续子序列中元素和最大的一个，
例如给定序列{ -2, 11, -4, 13, -5, -2 }，其最大连续子序列为{ 11, -4, 13 }，最大和
为20。
在今年的数据结构考卷中，要求编写程序得到最大和，现在增加一个要求，即还需要输出该
子序列的第一个和最后一个元素。

Input

测试输入包含若干测试用例，每个测试用例占2行，第1行给出正整数K( < 10000 )，第2行给出K个整数，中间用空格分隔。当K为0时，输入结束，该用例不被处理。

Output

对每个测试用例，在1行里输出最大和、最大连续子序列的第一个和最后一个元
素，中间用空格分隔。如果最大连续子序列不唯一，则输出序号i和j最小的那个（如输入样例的第2、3组）。若所有K个元素都是负数，则定义其最大和为0，输出整个序列的首尾元素。

Sample Input

6 -2 11 -4 13 -5 -2 10 -10 1 2 3 4 -5 -23 3 7 -21 6 5 -8 3 2 5 0 1 10 3 -1 -5 -2 3 -1 0 -2 0

Sample Output

20 11 13 10 1 4 10 3 5 10 10 10 0 -1 -2 0 0 0

求最大连续子序列难点在输出子序列的前后两个端点的值：

核心代码为：

for (int i = 2; i <= n; i++)
		{
			if (dp[i - 1] + a[i]>a[i])
			{
				dp[i] = dp[i - 1] + a[i];
			}
			else
			{
				dp[i] = a[i];
				temp = a[i];
			}
			if (Max < dp[i])
			{
				l = temp;
				r = a[i];
				Max = dp[i];
			}
		}

用temp保存开头的位置， 结尾的位置即为i，然后找到每次找到最大之后再用l 和 r 保存；

AC代码：

#include
#include
#include
#include
using namespace std;
#define Maxn 10005
int dp[Maxn];
int a[Maxn];
int main()
{
	int n;
	while (~scanf("%d", &n) && n)
	{
		int Max = 0;
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
		}
		dp[1] = a[1];
		Max = a[1];
		int l = a[1];
		int r = a[1];
		int temp = a[1];
		for (int i = 2; i <= n; i++)
		{
			if (dp[i - 1] + a[i]>a[i])
			{
				dp[i] = dp[i - 1] + a[i];
			}
			else
			{
				dp[i] = a[i];
				temp = a[i];
			}
			if (Max < dp[i])
			{
				l = temp;
				r = a[i];
				Max = dp[i];
			}
		}
		if (Max < 0) { printf("0 %d %d\n", a[1], a[n]); }
		else cout << Max << " " << l << " " << r << endl;
	}
}

#include
#include
#include
using namespace std;
#define Maxn 10005
int dp[Maxn];
int a[Maxn];
int main()
{
	int n;
	while (~scanf("%d", &n) && n)
	{
		int Max = 0;
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
		}
		dp[1] = a[1];
		Max = a[1];
		int l = a[1];
		int r = a[1];
		int temp = a[1];
		for (int i = 2; i <= n; i++)
		{
			if (dp[i - 1] + a[i]>a[i])
			{
				dp[i] = dp[i - 1] + a[i];
			}
			else
			{
				dp[i] = a[i];
				temp = a[i];
			}
			if (Max < dp[i])
			{
				l = temp;
				r = a[i];
				Max = dp[i];
			}
		}
		if (Max < 0) { printf("0 %d %d\n", a[1], a[n]); }
		else cout << Max << " " << l << " " << r << endl;
	}
}

HDU-1003也是类似的题目

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 247757 Accepted Submission(s): 58534

Problem Description

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

AC代码：

#include
#include
#include
#include
using namespace std;
#define Maxn 100005
int dp[Maxn];
int a[Maxn];
int main()
{
	int time = 1;
	int T;
	scanf("%d", &T);
	while (T--)
	{
		int n;
		scanf("%d", &n);
		int Max = 0;
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
		}
		dp[1] = a[1];
		Max = a[1];
		int l = 1;
		int r =  1;
		int temp = 1;
		for (int i = 2; i <= n; i++)
		{
			if (dp[i - 1] + a[i] >= a[i])
			{
				dp[i] = dp[i - 1] + a[i];
			}
			else
			{
				dp[i] = a[i];
				temp = i;
			}
			if (Max < dp[i])
			{
				l = temp;
				r = i;
				Max = dp[i];
			}
		}
		printf("Case %d:\n", time++);
		cout << Max << " " << l<< " " << r<< endl;
		if (T != 0)
			cout << endl;
	}
}

#include
#include
#include
using namespace std;
#define Maxn 100005
int dp[Maxn];
int a[Maxn];
int main()
{
	int time = 1;
	int T;
	scanf("%d", &T);
	while (T--)
	{
		int n;
		scanf("%d", &n);
		int Max = 0;
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
		}
		dp[1] = a[1];
		Max = a[1];
		int l = 1;
		int r =  1;
		int temp = 1;
		for (int i = 2; i <= n; i++)
		{
			if (dp[i - 1] + a[i] >= a[i])
			{
				dp[i] = dp[i - 1] + a[i];
			}
			else
			{
				dp[i] = a[i];
				temp = i;
			}
			if (Max < dp[i])
			{
				l = temp;
				r = i;
				Max = dp[i];
			}
		}
		printf("Case %d:\n", time++);
		cout << Max << " " << l<< " " << r<< endl;
		if (T != 0)
			cout << endl;
	}
}