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 n nodes but lacks of n − 1 edges. You
want to complete this tree by adding n − 1 edges. There must be exactly one path between any two
nodes after adding. As you know, there are n
n−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), where f is a predefined function and d is the degree of this node. What’s
the maximum coolness of the completed tree?
Input
The first line contains an integer T indicating the total number of test cases. Each test case starts with
an integer n in one line, then one line with n − 1 integers f(1), f(2), . . . , f(n − 1).
• 1 ≤ T ≤ 2015
• 2 ≤ n ≤ 2015
• 0 ≤ f(i) ≤ 10000
• There are at most 10 test cases with n > 100.
Output
For each test case, please output the maximum coolness of the completed tree in one line.
Sample Input
2
3
2 1
4
5 1 4
Sample Output
5
19

题目大概：

给你一个树的结点数量n，让你构造一棵树，使得这棵树的价值最大，每个结点都有一个价值，度数d，每个结点的价值为f（d）。这棵树的价值为所有结点的价值之和。f（1），f（2）....f（n-2），f（n-1）这些价值都会给出。

思路：

首先感觉这很像是一个背包问题，有一个固定的总度数2*n-2，每个度数都有一个价值，求最大价值max是多少。

但是每个结点有个限制，就是最少度数为1。那么我们可以预先为每个结点分1度，那么这时候树的价值为sum=n*f（1）。

这时候还有n-2度，然后需要计算出n-2度的背包的最大值。

这时候不能忘记我们已经给每个结点分配了1度的价值了，所以在背包中，每次添加一个物品的时候，需要把度为1的物品的价值给减去，防止多计算。

理清楚了所有内容，直接写背包就行了。

代码：

#include <bits/stdc++.h>

using namespace std;
const int INF=0x3f3f3f3f;
#define ll long long
const int maxn=2e4+10;
int dp[maxn];
int w[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<n;i++)
        {
            scanf("%d",&w[i]);
        }
        for(int i=0;i<=n;i++)dp[i]=-INF;
        int sum=w[1]*n;
        int m=n-2;
        dp[0]=0;//初始化
        for(int i=1;i<n-1;i++)
        {
            for(int j=i;j<=m;j++)
            {
                if(dp[j]<dp[j-i]+w[i+1]-w[1])
                {
                    dp[j]=dp[j-i]+w[i+1]-w[1];
                }
            }
        }
        printf("%d\n",dp[m]+sum);
    }
    return 0;
}