HDU 5534 Partial Tree（背包）

思路：一棵树有2(n-1)个度，每个度都有它的权值，那么就相当于一个容量为2*(n-1)的背包，物品的体积是度数，可是这样有可能会出现没有被选的度数，那么我们就先每个点都分配一个度，然后就是完全背包啦

#include<bits/stdc++.h>
using namespace std;
const int maxn = 25000;
int a[maxn];
int dp[maxn];
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i = 0;i<n-1;i++)
            scanf("%d",&a[i]);
        int V = 2*(n-1)-n;
		for(int i = 0;i<=n;i++)
			dp[i]=-1e9;
        dp[0]=a[0]*n;
        for(int i=1;i<n-1;i++)
            a[i]-=a[0];
		for(int i = 1;i<=V;i++)
			for(int j = i;j<=V;j++)
				dp[j]=max(dp[j],dp[j-i]+a[i]);
		printf("%d\n",dp[V]);
    }
    return 0;
}

Description

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 

 nodes but lacks of 

 edges. You want to complete this tree by adding 

 edges. There must be exactly one path between any two nodes after adding. As you know, there are 

 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 

, where 

 is a predefined function and 

Input

 indicating the total number of test cases. 

Each test case starts with an integer 

 in one line, 

then one line with 

 integers 

. 

 

 

 

There are at most 

 test cases with 

.

Output

For each test case, please output the maximum coolness of the completed tree in one line.

Sample Input

232 1 4 5 1 4

Sample Output

519