In mathematics, and more specifically in graph theory, a tree is an undirected graph in which any two nodes are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.

You find a partial tree on the way home. This tree has nn nodes but lacks of n−1n−1 edges. You want to complete this tree by adding n−1n−1edges. There must be exactly one path between any two nodes after adding. As you know, there are nn−2nn−2 ways to complete this tree, and you want to make the completed tree as cool as possible. The coolness of a tree is the sum of coolness of its nodes. The coolness of a node is f(d)f(d), where ff is a predefined function and dd is the degree of this node. What's the maximum coolness of the completed tree?

2

3

2 1

4

5 1 4

5

19

因为每种装的范围为【1，+无穷】，然而分组背包的三次方会T，所以要想办法将其转化为完全背包模型。
因为每个点都至少有一度，所以我们预先放入n个一度，本来要放的2×n-2度现在只剩n-2度。
每当再次放入一个x度时，他的贡献为x-1度，在放入的同时减掉之前预先放好的一度。这样便保证了每个点至少有一度。

#include<bits/stdc++.h>

#define MAX 2018

#define INF 0x3f3f3f3f

using namespace std;

typedef long long ll;

int a[MAX];

ll dp[MAX];

int main()

{

    int t,n,m,i,j,k;

    scanf("%d",&t);

    while(t--){

        scanf("%d",&n);

        for(i=;i<n;i++){

            scanf("%d",&a[i]);

        }

        memset(dp,-INF,sizeof(dp));

        dp[]=;

        for(i=;i<n;i++){

            for(j=i-;j<=n-;j++){

                dp[j]=max(dp[j],dp[j-(i-)]+a[i]-a[]);

            }

        }

        printf("%I64d\n",dp[n-]+n*a[]);

    }

    return ;

}