題目描述

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?

輸入

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.

輸出

For each test case, please output the maximum coolness of the completed tree in one line.

樣例輸入

2
3
2 1
4
5 1 4

樣例輸出

5
19

圖, 每個度 都有一個權值,

n-1條邊 去構造圖,   儘量滿足可以形成圖的情況下, 度數權值最大.求最大值.

題解:

很容易想到 揹包,   ,而且是不限量的揹包,  即完全揹包.  

一個圖的度數 是固定的 即 2*n-2;  又要保證圖存在, 那麼每個點至少度為1.  則   容量為 n-2

即 n個點分配n-2的度.

那麼完全揹包 可解:

初始值為 n*f[1]    揹包的時候  權值為 f[i+1] -f[1]  就可以/

[程式碼]

#include <bits/stdc++.h>
#include <stdio.h>
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=n;i>=a;i--)

typedef long long ll;
const int maxn = 1e5+10;
const int mod =1e9+7;
const int inf = 0x3f3f3f3f;
using namespace std;

int dp[maxn];
int f[maxn];
int main(int argc, char const *argv[])
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		memset(dp,-inf,sizeof(dp));
		memset(f,0,sizeof(f));
		int n;
		scanf("%d",&n);
		rep(i,1,n-1)
		{
			scanf("%d",&f[i]);
		}
		// V = 2*n-2 - n  =  n-2;
		dp[0] = n*f[1];
		rep(i,1,n-2)
		{
			rep(j,i,n-2)
			{
				dp[j] = max(dp[j],dp[j-i]+f[i+1]-f[1]);
			}
		}
		printf("%d\n",dp[n-2] );
	}
	return 0;
}