hdu 5534 (徹底揹包) Partial Tree

題目：這裏

題意：

感受並不能表達清楚題意，因此

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

 

首先，這個最終答案是與點的度有關，因爲是個樹，能夠知道最後全部點的度數和是n*2-2，還有，每一個點至少得有一個度，因此最終答案得先加上f[1]*n，而後如今

還剩下n-2個度，須要在n個點裏分配，使得分配以後的權值最大，可是這個分配因爲是有關聯的，一個點的度數加了1以後必須得有另外一個點的度數也加1，因此咱們的

分配方案還得知足這個條件，不能隨意分配，可是經過隨意取幾個n值構造一下樹發現，n-2個度任意分給n個點的方案可以知足構造出一棵樹，並且這個構造還挺有

規律，有遞推性，因此大膽認爲能夠任意分配，好，如今n-2個度分配給n個點，每次能夠分配1到n-1個度，問怎麼分配值f()最大，這不就是一個揹包麼，仍是一個徹底

揹包。再注意一下這是在每一個點已經有了一個度的前提下，因此得減去f[1]。

 

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 #define inf 0x3f3f3f3f
 8 const int M  = 1e4 + 10;
 9 int dp[M],a[M];
10 
11 int max(int x,int y){return x>y?x:y;}
12 
13 int main()
14 {
15     int t,n;
16     scanf("%d",&t);
17     while (t--){
18        scanf("%d",&n);
19        for (int i=1 ; i<n ; i++) {
20            scanf("%d",&a[i]);
21            if (i!=1) a[i]-=a[1];
22        }
23        //int pa=n*2-2;
24        for (int i=0 ; i<=n ; i++) dp[i]=-inf;
25        dp[0]=0;//dp[1]=a[1];
26        for (int i=2 ; i<n ; i++) {
27            for (int j=0 ; j+i-1<=n-2 ; j++)
28               dp[i+j-1] = max(dp[i+j-1],dp[j]+a[i]);
29        }
30        printf("%d\n",dp[n-2]+n*a[1]);
31     }
32     return 0;
33 }