9262: Partial Tree

時間限制: 1 Sec  記憶體限制: 128 MB 提交: 70  解決: 29 [提交] [狀態] [討論版] [命題人:admin]

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

[提交] [狀態]

比賽時看到這道題感覺是窒息的，但是賽後看題解更窒息，完全想不到是個揹包，極其震驚

題意：給n各點，分配n-1條邊，使每個點的度i對應的f（ i ）相加的和最大，f（i）給出

思路：大佬說，不管怎麼建樹，只要能夠保證每個點的度至少為一，那麼總是存在一棵樹滿足你的度數分配要求，那麼只需要把

度數分配下去，並且每個點的度數至少為1，所以先給每個點都分配1的度數，所以最初的2n-2的度數就剩下了n-2的度數。

然後就可以看作完全揹包的問題，把n-2總度數看作總空間的體積，每個i為體積，f（i）為對應的權值，然後在滿足總體積的前提

下讓權值最大

程式碼：

#include <bits/stdc++.h>

using namespace std;
const int mod = 1e9 + 7;
const int maxn = 1e4+10;
const int inf = 0x3f3f3f3f;
typedef long long ll;
ll f[maxn],dp[maxn];
int main() {
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for(int i=1; i<n; i++){
            scanf("%lld",&f[i]);
            dp[i]=-inf;
        }
        dp[n]=-inf;
        dp[0]=f[1]*n;
        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("%lld\n",dp[n-2]);
    }
    return 0;
}