Partial Tree(DP)

Partial Tree

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

Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)

Total Submission(s):     Accepted Submission(s):

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− edges. You want to complete this tree by adding n− edges. There must be exactly one path between any two nodes after adding. As you know, there are nn− 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− integers f(),f(),…,f(n−).

≤T≤

≤n≤

≤f(i)≤

There are at most  test cases with n>.

Output

For each test case, please output the maximum coolness of the completed tree in one line.

Sample Input

Sample Output

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[];

 int dp[];

 int main(){

     int t;

     scanf("%d",&t);

     while(t--){

         int n;

         scanf("%d",&n);

         for(int i=;i<n;i++){

             scanf("%d",&a[i]);

         }

         int ans=a[]*n;

         for(int i=;i<=n;i++) dp[i]=-INF;

         dp[]=;

         for(int i=;i<n;i++) a[i]-=a[];

         for(int i=;i<n;i++){

             for(int j=;j<=n-;j++){

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

             }

         }

         printf("%d\n",ans+dp[n-]);

     }

 }