HDU 5534 - Partial Tree（完全背包）【真*好题】

Partial Tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 2087    Accepted Submission(s): 1035   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）     Recommend hujie   |   We have carefully selected several similar problems for you:  6447 6446 6445 6444 6443      Statistic | Submit | Discuss | Note

 

思路：刚开始一看就以为是树形DP，想着凉了。后来仔细思考了一下，看着像是背包。

对于一张图G，n个结点，n-1条边，则各个结点的度的和 sum = (n-1) * 2；这里，sum 就相当于背包的容量 V 。各个点的度和权值就相当于物品的体积和价值。我们这样就把问题转化成了一个完全背包。

但是，如果直接变成一个完全背包的话是错误的，因为有些结点可能没有选（这样就不能构成一棵树了）。

所以我们进行一波骚操作：

把所有点的度初始都设定为a[1]，初始的权值和为 a[1]*n（即dp[0]）。

这样，我们之后只要使得度数之和变成（n-2）即可。

dp[i]表示度数为 i 条件下的最大收益。

状态转移方程：dp[i] = max( dp[i-k] + a[k+1] - a[1] )。意思是，我们想要更新度数为i的最大权值，我们会想着它可以基于之前的一个度数，然后把一个未扩展的子节点扩展a[k+1]-a[1]的权值。

1. 我们有n个点可以扩展度数，而可能扩展的总度数只有n-2，所以一定不会出现多次扩展的情况

2. 每个点都有一个为1的基准度，一定有办法把它们恰好拼成一棵树。

 

 

#pragma GCC optimize(2)
#include 
using namespace std;
#define clr(a) memset(a,0,sizeof(a))
#define line cout<<"-----------------"<