hdu--5534

Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 578    Accepted Submission(s): 286

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）

 

题目大意：告诉你有n各结点，f（i）表示度为i的结点的cool值，现在你需要做的就是加n-1条边，构成一棵树，并使得这棵树的每个结点的cool值之和，即coolness最大，输出最大的coolness。

解体思路：这题如果不看人家写的，我想不到用完全背包。是一道好题，值得好好研究研究。

一棵树，n个结点，则度的总和为2*（n-1）.每一个结点至少有1个度,则首先给每个结点一个度，剩下2*（n-1）-n就在这n个结点中随机分配，得到一个cooldness最大的分配方式。可以看成完全背包问题

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int f[2020];
int dp[2020];//dp[i]表示增加i个度所得到的coolness值 
int main(){
	int t,n,v,w,i,j;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
	    for(i=1;i<n;i++){
	    	scanf("%d",&f[i]);
	    }
	    v=2*(n-1)-n;//每一个结点至少有1个结点,最多有1+n-2个结点 
	    memset(dp,0,sizeof(dp));
		dp[0]=n*f[1];//初始化 ,即一开始为每个结点只有一个度 
		for(i=2;i<n;i++){
			for(j=i-1;j<=v;j++)//v相当于背包容量，每一i物品的花费是i-1，价值是f[i]-f[1]
	    	dp[j]=max(dp[j],dp[j-i+1]+f[i]-f[1]);//减去f[1],因为每一个点已经度为1了 
	    }
	    printf("%d\n",dp[v]);
	}
	return 0;
}