hdu5534 Partial Tree

Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 599    Accepted Submission(s): 294

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 arenn−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 isf(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

232 145 1 4

 

Sample Output

519

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

题意：构建一个n个点的树，每个点的度i的价值为f(i)，求能获得的最大价值。

思路：树的度总和是2*（n-1）,把题目看作是将2*（n-1）个物品放进n个背包，看作背包问题，但是这样还有个条件限制就是每个背包至少要有1个物品（1个度），这样做要用O（n^3），但如果事先每个点分配一个度，那么其他的度就可以任意分配了，现在题目可以看作，把n-2个物品放进1个背包里，并且f(i)都要减去f(1)。

#include <stdio.h>#include <string.h>#define maxn 2050#define ll long long intconst int inf = 0x3f3f3f3f;ll w[maxn], dp[maxn];ll max(ll a, ll b){    return a>b?a:b;}int main(){    ll T, n, i, j, v;    scanf("%I64d", &T);    while(T--){        scanf("%I64d", &n);        for(i = 1;i < n;i++)            scanf("%I64d", &w[i]);        memset(dp, 0, sizeof dp);        dp[0] = n*w[1];//先每个点分配一个度        v = 2*(n - 1) - n;//现在的每个点最多分配n - 2个度        for(i = 2;i < n;i++)            for(j = i - 1;j <= v;j++)//原先已经分配了一个度，要减去1                dp[j] = max(dp[j], dp[j - i + 1] + w[i] - w[1]);//贡献值都要减去w[1]        printf("%I64d\n", dp[v]);    }}