CF EDU 108 D - Maximum Sum of Products

D - Maximum Sum of Products

区间DP

可先算出不反转任何子区间的答案 \(sum\)

再利用 区间DP 算出反转 \([l,r]\) 的话会比原来的答案多多少，记为 \(f[l][r]\)

转移：\(f[l][r]=f[l+1][r-1]+a[l]*b[r]+a[r]*b[l]-a[l]*b[l]-a[r]*b[r]\)

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;
typedef long long ll;
const int N = 5e3 + 10;
int n;
ll f[N][N];
ll a[N], b[N], sum;
int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> a[i];
	for (int i = 1; i <= n; i++) cin >> b[i];
	for (int i = 1; i <= n; i++) sum += a[i] * b[i];
	for (int i = 1; i < n; i++) f[i][i+1] = a[i] * b[i+1] + a[i+1] * b[i] - a[i] * b[i] - a[i+1] * b[i+1];
	for (int len = 3; len <= n; len++)
	{
		for (int l = 1; l + len - 1 <= n; l++)
		{
			int r = l + len - 1;
			f[l][r] = f[l+1][r-1] + a[l] * b[r] + a[r] * b[l] - a[l] * b[l] - a[r] * b[r];
		}
	}
	ll add = 0;
	for (int len = 2; len <= n; len++)
	{
		for (int l = 1; l + len - 1 <= n; l++)
		{
			int r = l + len - 1;
			add = max(add, f[l][r]);
		}
	}
	cout << sum + add << endl;
	return 0;
}