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

Tabris

CSDNhdu动态规划

发布于：2017年7月5日

次浏览

[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

版权声明：本文为 Tabris 原创文章，未经博主允许不得私自转载。
https://blog.csdn.net/qq_33184171/article/details/74374144

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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

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;
}

本页文档最后更新于：2021年5月16日

CSDN

动态规划

思维