Maximum Sum of Products

题面翻译

给定两个长为 $n$ 的序列 $a$ 和 $b$。你可以对 $a$ 的一段区间翻转，也可以不翻转，要求翻转后 $a$ 与 $b$ 对应位置之积的和最大。即求下式的值最大：
$$\sum_{i=1}^na_i\times b_i$$

其中 $n\le 5000$，$a_i,b_i\le10^7$。

translated by @zc_Ethan

题目描述

You are given two integer arrays $ a $ and $ b $ of length $ n $ .

You can reverse at most one subarray (continuous subsegment) of the array $ a $ .

Your task is to reverse such a subarray that the sum $ \sum\limits_{i=1}^n a_i \cdot b_i $ is maximized.

输入格式

The first line contains one integer $ n $ ( $ 1 \le n \le 5000 $ ).

The second line contains $ n $ integers $ a_1, a_2, \dots, a_n $ ( $ 1 \le a_i \le 10^7 $ ).

The third line contains $ n $ integers $ b_1, b_2, \dots, b_n $ ( $ 1 \le b_i \le 10^7 $ ).

输出格式

Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of $ a $ .

样例 #1

样例输入 #1

5
2 3 2 1 3
1 3 2 4 2

样例输出 #1

29

样例 #2

样例输入 #2

2
13 37
2 4

样例输出 #2

174

样例 #3

样例输入 #3

6
1 8 7 6 3 6
5 9 6 8 8 6

样例输出 #3

235

提示

In the first example, you can reverse the subarray $ [4, 5] $ . Then $ a = [2, 3, 2, 3, 1] $ and $ 2 \cdot 1 + 3 \cdot 3 + 2 \cdot 2 + 3 \cdot 4 + 1 \cdot 2 = 29 $ .

In the second example, you don’t need to use the reverse operation. $ 13 \cdot 2 + 37 \cdot 4 = 174 $ .

In the third example, you can reverse the subarray $ [3, 5] $ . Then $ a = [1, 8, 3, 6, 7, 6] $ and $ 1 \cdot 5 + 8 \cdot 9 + 3 \cdot 6 + 6 \cdot 8 + 7 \cdot 8 + 6 \cdot 6 = 235 $ .

题目范围小，考虑暴力，可以枚举每一个区间，要发现如果交换ai和aj，那么对答案的贡献就是(ajbi+aibj)-(aibi+ajbj)即(ai-aj)*(bj-bi)，所以应该枚举每一个点当作区间中点，左右两边扩张区间，找到最大的增量区间，但是区间中点不一定有对称的与中间的值，所以l=i，r=i和l=i，r=i+1都要试一遍

const int N = 1e6 + 10;
int a[N];
int b[N];

void solve()
{
    int n; cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) cin >> b[i];
    int mmax = 0;
    for (int i = 1; i <= n; i++) mmax += a[i] * b[i];

    int jia = 0;
    for (int i = 1; i <= n; i++)
    {
        int l = i, r = i;
        int ans = 0;
        while (l >= 1 && r <= n)
        {
            ans += (a[l] - a[r]) * (b[r] - b[l]);
            jia = max(jia, ans);
            l--, r++;
        }
    }
    for (int i = 1; i <= n - 1; i++)
    {
        int l = i, r = i+1;
        int ans = 0;
        while (l >= 1 && r <= n)
        {
            ans += (a[l] - a[r]) * (b[r] - b[l]);
            jia = max(jia, ans);
            l--, r++;
        }
    }
    cout << mmax+jia << '\n';
}