HDU 5534 Partial Tree（动态规划）

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

题面：

Partial Tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 763    Accepted Submission(s): 374


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  
Source 2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）  

题意：

    给定一颗无向树的节点数和各个度对应的权值，定义酷度为每个节点的度数乘以相应的度对应的权值的总和，求最大酷度。

解题：

    比较直接的想法是dp[i][j]，表示前i个节点分配了j点度数能获取的最大值，写法清晰明了，可惜复杂度为n^3，明显超时。

    结合树的特殊结构，每个节点的度数最少为1，因此预先给每个节点分配一点度数，随后再将剩余的2*(n-1)-n点度数分配，使之得到最大的酷度。

    状态转移方程可以改进为dp[i]，代表在原来均分了n点度数的基础上，再分配i点度数所能获取的最大酷度。因为已经保障了每个节点至少为1度，后续分配也不会出现多次替换的情况，因此省却了位置这一维，复杂度降为n^2.问题得解。

    dp[i]=max(dp[i],dp[i-j]+f[j+1]-f[1])；

代码：

#include
#include
#include
#include
#include 
using namespace std;
int f[2020],dp[2020];
int main()
{
	int t,n,lim;
	scanf("%d",&t);
	while(t--)
	{
		memset(dp,0,sizeof(dp));
		scanf("%d",&n);
		for(int i=1;i