hdu 5534 Partial Tree(完全背包)

 

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）

 

注意到一个节点数为n的树的度数和为2*n-2,所以问题就转换为了把2*n-2个度分配给n个节点所能获得的最大价值，而且每一个节点至少分到1个度。我们可以先每一个分一个度，然后把n-2个节点任意分配完。分配的时候因为已经分了1个度了，所以要把2~n-1的度看为1~n-1，然后做个完全背包就行了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 2116
#define inf 1e12
int n;
int f[N],dp[N];
int main()
{
   int t;
   scanf("%d",&t);
   while(t--){
      scanf("%d",&n);
      for(int i=1;i<=n-1;i++){
         scanf("%d",&f[i]);
      }
      int ans=0;
      ans+=f[1]*n;
      for(int i=2;i<=n-1;i++){
         f[i]-=f[1];
      }
      for(int i=1;i<=n-2;i++){
         f[i]=f[i+1];
      }
      //memset(dp,0,sizeof(dp));
      for(int i=0;i<N;i++){
         dp[i]=-inf;
      }
      dp[0]=0;
      for(int i=1;i<=n-2;i++){
         for(int j=i;j<=n-2;j++){
            dp[j]=max(dp[j],dp[j-i]+f[i]);
         }
      }
      ans+=dp[n-2];
      printf("%d
",ans);
   }
   return 0;
}