HDU 5534 Partial Tree（动态规划）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5534

题面：

Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 763    Accepted Submission(s): 374

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：

    给定一颗无向树的节点数和各个度对应的权值，定义酷度为每个节点的度数乘以相应的度对应的权值的总和，求最大酷度。

解题：

    比较直接的想法是dp[i][j]，表示前i个节点分配了j点度数能获取的最大值，写法清晰明了，可惜复杂度为n^3，明显超时。

    结合树的特殊结构，每个节点的度数最少为1，因此预先给每个节点分配一点度数，随后再将剩余的2*(n-1)-n点度数分配，使之得到最大的酷度。

    状态转移方程可以改进为dp[i]，代表在原来均分了n点度数的基础上，再分配i点度数所能获取的最大酷度。因为已经保障了每个节点至少为1度，后续分配也不会出现多次替换的情况，因此省却了位置这一维，复杂度降为n^2.问题得解。

    dp[i]=max(dp[i],dp[i-j]+f[j+1]-f[1])；

代码：

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include <algorithm>using namespace std;int f[2020],dp[2020];int main(){int t,n,lim;scanf("%d",&t);while(t--){memset(dp,0,sizeof(dp));scanf("%d",&n);for(int i=1;i<n;i++)scanf("%d",&f[i]);dp[0]=n*f[1];for(int i=1;i<=n-2;i++)for(int k=1;k<=i;k++)dp[i]=max(dp[i],dp[i-k]+f[k+1]-f[1]);printf("%d\n",dp[n-2]);}return 0;}