HDU-5534-Partial Tree

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]);
    }
}