G. Game With Triangles: Season 2【Codeforces Round 1009 (Div. 3)】

G. Game With Triangles: Season 2

思路：

很模板的区间dp
$\mathcal{O}(n^3)$，转移方程如下：
$${\text{dp}}_{[L,R]}=\max \left(\underset{L<i<R}{\max }\left({\text{dp}}_{[L+1,i-1]}+{\text{dp}}_{[i+1,R-1]}+\text{score}(L,i,R)\right),\underset{L\le i<R}{\max }\left({\text{dp}}_{[L,i]}+{\text{dp}}_{[i+1,R]}\right)\right)$$

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n'
#define int long long
#define pb push_back
#define pii pair<int, int>
#define FU(i, a, b) for (int i = (a); i <= (b); ++i)
#define FD(i, a, b) for (int i = (a); i >= (b); --i)
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int n;
int a[405];
int dp[405][405]; // L-R区间上的最大val值
int val(int a, int b, int c) { return a * b * c; }
int modn(int x) { return (x + n - 1) % n + 1; }

void solve() {
    cin >> n;
    FU(i, 1, n) { cin >> a[i]; }
    FU(len, 1, n) {                    // 枚举长度
        for (int l = 1; l <= n; l++) { // 枚举左端点
            int ans = 0;
            int r = modn(l + len - 1); // 枚举右端点
            if (len <= 2) {            // 长度小于等于2 为0
                dp[l][r] = 0;
                continue;
            }
            if (len == 3) {
                dp[l][r] = val(a[l], a[modn(l + 1)], a[r]);
                continue;
            }
            for (int i = modn(l + 1); i != r;
                 i = modn(i + 1)) { // 枚举L+1至R-1上的点i，考虑三角形LiR
                int tmp = val(a[l], a[i], a[r]);
                if (i != modn(l + 1))
                    tmp += dp[modn(l + 1)][modn(i - 1)];
                if (i != modn(r - 1))
                    tmp += dp[modn(i + 1)][modn(r - 1)];
                ans = max(ans, tmp);
            }
            for (int i = l; i != r; i = modn(i + 1)) { // 枚举分为两段区间合
                ans = max(ans, dp[l][i] + dp[modn(i + 1)][r]);
            }
            dp[l][r] = ans;
        }
    }
    int res = 0;
    for (int i = 1; i <= n; i++) {
        res = max(res, dp[i][modn(i + n - 1)]);
    }
    cout << res << endl;
}

signed main() {
    cin.tie(0)->ios::sync_with_stdio(0);
    int T = 1;
    cin >> T;
    while (T--) {
        solve();
    }
    return 0;
}