CF1899C Yarik and Array（DP，贪心）

题目链接

题目

A subarray is a continuous part of array.

Yarik recently found an array a
of n
elements and became very interested in finding the maximum sum of a non empty subarray. However, Yarik doesn’t like consecutive integers with the same parity, so the subarray he chooses must have alternating parities for adjacent elements.

For example, [1,2,3]
is acceptable, but [1,2,4]
is not, as 2
and 4
are both even and adjacent.

You need to help Yarik by finding the maximum sum of such a subarray.

Input
The first line contains an integer t
(1≤t≤104)
— number of test cases. Each test case is described as follows.

The first line of each test case contains an integer n
(1≤n≤2⋅105)
— length of the array.

The second line of each test case contains n
integers a1,a2,…,an
(−103≤ai≤103)
— elements of the array.

It is guaranteed that the sum of n
for all test cases does not exceed 2⋅105
.

Output
For each test case, output a single integer — the answer to the problem.

Example

inputCopy
7
5
1 2 3 4 5
4
9 9 8 8
6
-1 4 -1 0 5 -4
4
-1 2 4 -3
1
-1000
3
101 -99 101
20
-10 5 -8 10 6 -10 7 9 -2 -6 7 2 -4 6 -1 7 -6 -7 4 1

outputCopy
15
17
8
4
-1000
101
10

题目大意

$t$ 组测试数据
每组给一个整数 $n$ 和 $n$ 个整数（包含负数），问在这 $n$ 个整数中，找和最大的连续子序列，要求是该子序列里的数是 奇偶交替的。

思路

这题我用的是dp的思路，定义一个状态数组f，对于 $f_i$ 是以 $a_i$ 开头的数列的最大值，
从后往前开始推，因为每个$f_i$都是最大的，$f_{i-1}$ 只用考虑在符合要求的前提下，是否要接上后面那串数列了。
$f_i=max(f_i,f_i+f_{i+1})$

代码

#include
using namespace std;
const int N =1e6 + 10;
int a[N], f[N];
int main()
{
	int T; cin >> T;
	while (T -- )
	{
		int n; scanf("%d", &n);
		for (int i = 1; i <= n; i ++ )
		{
			scanf("%d", &a[i]);
			f[i] = a[i]; //初始化状态 
		}
		int ans = -0x3f3f3f3f;
		
		for (int i = n - 1; i >= 1; i -- )
		{
//			判断条件，是否符合一奇一偶的顺序 
			if (abs(a[i]) % 2 != abs(a[i + 1]) % 2)
			{
				f[i] = max(f[i], f[i] + f[i + 1]);
			}
		}
		for (int i = 1; i <= n; i ++ )
		{
			ans = max(ans, f[i]);
		}
		printf("%d\n", ans);
	}
	return 0;
}

总结

好久没打CF，生疏了，第二题暴力，思路对了，但很多细节没写好，比如最大值答案用了0x3f3f3f3f，事实上在long long int 的情况下0x3f3f3f3f是不够大的，连续改大了两次才写对。

这题先写了一遍暴力，错了，然后才想的优化。
暴力版

#include
using namespace std;
const int N =1e6 + 10;
int a[N], f[N];
int main()
{
	int T; cin >> T;
	while (T -- )
	{
		int n; scanf("%d", &n);
		for (int i = 1; i <= n; i ++ )
		{
			scanf("%d", &a[i]);
			f[i] = a[i]
		}
		int ans = -0x3f3f3f3f;
		for (int i = 1; i <= n; i ++ )
		{
			ans = max(ans, f[i]);
		}
		for (int i = 1; i <= n; i ++ )
		{
			int sum = a[i];
			ans = max(sum, ans);
			for (int j = i + 1; j <= n; j ++ )
			{
//				cout << a[j] << " " << a[j] % 2 << endl;
//				cout << a[j  - 1] << " " << a[j - 1] % 2 << endl;
				 
				if (abs(a[j] % 2) == abs(a[j - 1]) % 2)
				{
//					cout << 11111111 << endl;
					break;
				}
				sum += a[j];
				ans = max(ans, sum);
//				cout << ans << endl;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}

之前还写岔了一次，如果都是正数的话，可以像下面这么写

//正数版 
#include
using namespace std;
const int N =1e6 + 10;
int a[N];
int main()
{
	int T; cin >> T;
	while (T -- )
	{
		int n; cin >> n;
		for (int i = 1; i <= n; i ++ )
		{
			cin >> a[i];
		}
		int ans = -0x3f3f3f3f;
		int sum = a[1];
		ans = max(sum, ans);
		for (int i = 2; i <= n; i ++ )
		{
			if (a[i] % 2 == a[i - 1] % 2)
			{
				ans = max(sum, ans);
				sum = a[i];
			}
			else
			{
				sum += a[i];
			}
			ans = max(sum, ans);
		}
		ans = max(sum, ans);
		cout << "      " << ans << endl;
	}
	return 0;
}