hdu 5534 (完全背包) Partial Tree

题目：这里

题意：

感觉并不能表达清楚题意，所以

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，这个最终答案是与点的度有关，由于是个树，可以知道最后所有点的度数和是n*2-2，还有，每个点至少得有一个度，所以最终答案得先加上f[1]*n，然后现在

还剩下n-2个度，需要在n个点里分配，使得分配之后的权值最大，但是这个分配由于是有关联的，一个点的度数加了1之后必须得有另一个点的度数也加1，所以我们的

分配方案还得满足这个条件，不能随意分配，但是通过随意取几个n值构造一下树发现，n-2个度任意分给n个点的方案能够满足构造出一棵树，而且这个构造还挺有

规律，有递推性，所以大胆认为可以任意分配，好，现在n-2个度分配给n个点，每次可以分配1到n-1个度，问怎么分配值f()最大，这不就是一个背包么，还是一个完全

背包。再注意一下这是在每个点已经有了一个度的前提下，所以得减去f[1]。

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

#define inf 0x3f3f3f3f
const int M  = 1e4 + 10;
int dp[M],a[M];

int max(int x,int y){return x>y?x:y;}

int main()
{
    int t,n;
    scanf("%d",&t);
    while (t--){
       scanf("%d",&n);
       for (int i=1 ; i<n ; i++) {
           scanf("%d",&a[i]);
           if (i!=1) a[i]-=a[1];
       }
       //int pa=n*2-2;
       for (int i=0 ; i<=n ; i++) dp[i]=-inf;
       dp[0]=0;//dp[1]=a[1];
       for (int i=2 ; i<n ; i++) {
           for (int j=0 ; j+i-1<=n-2 ; j++)
              dp[i+j-1] = max(dp[i+j-1],dp[j]+a[i]);
       }
       printf("%d
",dp[n-2]+n*a[1]);
    }
    return 0;
}