Partial Tree(DP)

 Partial Tree

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

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2190    Accepted Submission(s): 1096


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 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?
 

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

2
3
2 1
4
5 1 4

 

Sample Output

5
19

 

Source
2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）
 

Recommend
hujie

 因为每个点最少的度数为1，所以可以先设每个点的度数为1，剩余的度数为(2*n-2)-n=n-2,相当于把剩下的n-2个度分给n个顶点，这样就转化为完全背包的题目了

#include<bits/stdc++.h>
const int INF=0x3f3f3f3f;
using namespace std;

int a[2505];
int dp[5505];

int main(){

    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for(int i=1;i<n;i++){
            scanf("%d",&a[i]);
        }
        int ans=a[1]*n;
        for(int i=1;i<=n;i++) dp[i]=-INF;
        dp[0]=0;
        for(int i=2;i<n;i++) a[i]-=a[1];
        for(int i=2;i<n;i++){
            for(int j=0;j<=n-2;j++){
                dp[i+j-1]=max(dp[i+j-1],dp[j]+a[i]);
            }
        }
        printf("%d
",ans+dp[n-2]);
    }

}