CF1389B-Array Walk

题目:

题目描述:

You are given an array $ a_1, a_2, \dots, a_n $ , consisting of $ n $ positive integers.

Initially you are standing at index $ 1 $ and have a score equal to $ a_1 $ . You can perform two kinds of moves:

1. move right — go from your current index $ x $ to $ x+1 $ and add $ a_{x+1} $ to your score. This move can only be performed if $ x<n $ .
2. move left — go from your current index $ x $ to $ x-1 $ and add $ a_{x-1} $ to your score. This move can only be performed if $ x>1 $ . Also, you can’t perform two or more moves to the left in a row.

You want to perform exactly $ k $ moves. Also, there should be no more than $ z $ moves to the left among them.

What is the maximum score you can achieve?

输入格式:

The first line contains a single integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of testcases.

The first line of each testcase contains three integers $ n, k $ and $ z $ ( $ 2 \le n \le 10^5 $ , $ 1 \le k \le n - 1 $ , $ 0 \le z \le min(5, k) $ ) — the number of elements in the array, the total number of moves you should perform and the maximum number of moves to the left you can perform.

The second line of each testcase contains $ n $ integers $ a_1, a_2, \dots, a_n $ ( $ 1 \le a_i \le 10^4 $ ) — the given array.

The sum of $ n $ over all testcases does not exceed $ 3 \cdot 10^5 $ .

输出格式:

Print $ t $ integers — for each testcase output the maximum score you can achieve if you make exactly $ k $ moves in total, no more than $ z $ of them are to the left and there are no two or more moves to the left in a row.

样例:

样例输入 1:

1
2
3
4
5
6
7
8
9

4
5 4 0
1 5 4 3 2
5 4 1
1 5 4 3 2
5 4 4
10 20 30 40 50
10 7 3
4 6 8 2 9 9 7 4 10 9

样例输出 1:

1
2
3
4

15
19
150
56

思路:

实现:

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10
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12
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19
20
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#include "ybwhead/ios.h"
int TTT;
int n, k, z;
const int maxn = 1e5 + 10;
int ans;
int a[maxn];
int dp[maxn][10];
int main()
{
    yin >> TTT;
    while (TTT--)
    {
        yin >> n >> k >> z;
        ++k;
        ans = 0;
        for (int i = 1; i <= n; i++)
        {
            yin >> a[i];
        }
        for (int i = 1; i <= k; i++)
        {
            for (int j = 0; j <= z; j++)
            {
                dp[i][j] = dp[i - 1][j] + a[i];
                if (!j)
                    ans = max(ans, dp[i][j]);
                if (j * 2 + i <= k && j)
                {
                    dp[i][j] = max(dp[i][j], dp[i][j - 1] + a[i] + a[i - 1]);
                    ans = max(ans, dp[i][j]);
                }
                if (j * 2 + i - 1 == k && j)
                {
                    // dp[i][j] = max(dp[i][j], dp[i][j - 1] + a[i - 1]);
                    ans = max(ans, dp[i][j - 1] + a[i - 1]);
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
}