[原创]HDU 5534 Partial Tree [完全背包]【动态规划+思维】

2017-07-05 09:31:07 Tabris_ 阅读数：480

博客爬取于2020-06-14 22:39:47
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/74374144

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

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1401 Accepted Submission(s): 693

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
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题目大意:
有n个节点,让你加n-1条边构建成一棵联通树,使得其权值和最大
题目给定节点度数为i的权值为f(i),

解题思路:

首先能够确定的是对于整棵树度数和为 $2*(n-1)$ ,
如果树联通的话,那么每个节点度数至少为 $1$ .

于是问题就变成如何分配 $2*(n-1)-n*1 = n-2$ 这几个度数

可以将这个问题变成在容量为 $n-2$ 的背包中不限次数拿取体积为 $i$ 价值为 $f(i)$ 这样一个完全背包问题.

之后就是因为节点度数至少为 $1$ 了,而在转移的时候至少都是度数为 $2$ 的节点

那么转移过程应该怎么办?

这样考虑,如果多了一个度数为 $x$ 的节点,那么同时也就少了一个度数为 $1$ 的节点

显然在转移的时候先减去 $f[1]$ ,最后在结果加回 $n*f[1]$ ,那么就把应该有的度数为 $1$ 的节点计算上了.又因为最后这个 $n*f[1]$ 是常数,不会影响dp转移的结果

附本题代码
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int n;

int f[2222];
int dp[2222];

int main(){
    for(int _=read();_;_--){
        n=read();
        for(int i=1;i<n;i++) f[i]=read(),dp[i]=-1e9; dp[0]=0;
        for(int i=1;i<n-1;i++)
            for(int j=i;j<=n-2;j++)
                dp[j]=max(dp[j],dp[j-i]+f[i+1]-f[1]);
        printf("%d\n",dp[n-2]+n*f[1]);
    }
    return 0;
}