HDU-5534-Partial Tree

Posted 2021-01-02 smallhester

Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2061    Accepted Submission(s): 1023

Problem 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

 

题意：给你一颗含有n个点的树，有n-1条边，每一个度数都有相应的value，问这棵树的最大的value值是多少

思路：n-1条边，也就是度数只和为2n-2，就可以转化为一个完全背包问题，一开始先把所有的点都看作是单个的一度顶点，之后在进行dp

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<set>
#include<vector>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
#define ll long long
int const maxn = 1e5+7;
const int mod = 1e9 + 7;
int gcd(int a, int b) {
    if (b == 0) return a;  return gcd(b, a % b);
}
int val[2020];
int dp[2020];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n-1;i++)
            scanf("%d",&val[i]);
        int v=2*(n-1)-n;      //一开始将所有的点都看作一度顶点，那么剩余的定点就是2*（n-1）-n，接下来将v个度数进行dp
        for(int i=1;i<=n-1;i++)
            dp[i]=-INF;
        dp[0]=val[1]*n;    //一开始将dp[0]初始化为所有的点都是一度顶点的总value值
        for(int i=2;i<=n-1;i++)     //从二度顶点到n-1度顶点开始循环dp
            for(int j=i-1;j<=v;j++)    //最多只有v个顶点可以进行调控
                dp[j]=max(dp[j],dp[j-i+1]+val[i]-val[1]);     //增添一个度数为i的顶点，一开始由于所有的顶点都是一度顶点，所以使用的cost是i-1，而得到的value值是val[i]-val[1]
        printf("%d
",dp[v]);
    }
}