9262: Partial Tree

時間限制: 1 Sec  記憶體限制: 128 MB
提交: 35  解決: 14
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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

來源/分類

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【題意】

給出一個數列a[]，讓我們構造一棵n點的樹，點i的度為x，則點i的價值為a[x]，求最大的n個點價值之和。

【分析】

容易知道，一棵樹總共有2n-2個度，而且這棵樹每個點必須至少有一個度，那我們就先給每個點一個度。剩下n-2個度可以自由分配。用動態規劃去分配這n-2個度，求出最大值。

設dp[i]表示分配了i個度的最優值。初始狀態dp[0]表示已經給每個點分配了一個度，故dp[0]=n*a[1];  

我們的揹包容量就是n-2，n-2個度分配給n個點，可能是給某個點分1度、2度或者i度，為某個點分配上i個度，則其度從1變為i+1。  分配i個度可能分配多次，故完全揹包可解。dp[j]=max(dp[j-i]+a[i+1]-a[1]);

【程式碼】

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

ll dp[2020];
int a[2020],n,T;
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(int i=1;i<n;i++)scanf("%d",&a[i]);
		for(int i=0;i<=n-2;i++)dp[i]=-1e9;
		dp[0]=n*a[1];
		for(int i=1;i<=n-2;i++)
		{
			for(int j=i;j<=n-2;j++)
				dp[j]=max(dp[j],dp[j-i]+a[i+1]-a[1]);
		}
		printf("%lld\n",dp[n-2]);
	}
}