Partial Tree

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5534

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 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 nn−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

HINT

题意

给你n个点，让你构造出一棵树

假设这棵树最后度数为k的点有num[k]个，那么这棵树的价值为sigma(num[i]*f[i])

其中f[i]是已经给定的

题解：

dp，我们首先给所有点都分配一个度数，那么还剩下n-2个度数没有分配

我们就可以dp了

dp[i]表示当前分配了i点度数时获得的最优值是多少，那么直接暴力转移就好了

背包DP

代码

#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
int n;
int dp[2300];
int f[2300];
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n-1;i++)
            scanf("%d",&f[i]);
        for(int i=0;i<=n;i++)
            dp[i]=-9999999;
        dp[0]=n*f[0];
        for(int i=1;i<n-1;i++)
            f[i]-=f[0];
        n-=2;
        for(int i=1;i<=n;i++)
        {
            for(int j=i;j<=n;j++)
            {
                dp[j]=max(dp[j],dp[j-i]+f[i]);
            }
        }
        printf("%d\n",dp[n]);
    }
}