hdu5534 Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 126    Accepted Submission(s): 68

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

232 145 1 4

 

Sample Output

519

 

注意到一个节点数为n的树的度数和玮2*n-2,所以问题就转换为了把2*n-2个度分配给n个节点所能获得的最大价值，而且每一个节点至少分到1个度。我们可以先每一个分一个度，然后把n-2个节点任意分配完。分配的时候因为已经分了1个度了，所以要把2~n-1的度看为1~n-1，然后做个完全背包就行了。

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;typedef long long ll;#define inf 99999999int v[2200],dp[2200];int main(){    int n,m,i,j,T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(i=1;i<=n-1;i++){            scanf("%d",&v[i]);        }        int ans=0;        ans+=v[1]*n;        for(i=2;i<=n-1;i++){            v[i]-=v[1];        }        for(i=1;i<=n-2;i++){            dp[i]=-inf;        }        dp[0]=0;        for(i=1;i<=n-2;i++){            v[i]=v[i+1];        }        for(i=1;i<=n-2;i++){            for(j=i;j<=n-2;j++){                dp[j]=max(dp[j],dp[j-i]+v[i]);            }        }        ans+=dp[n-2];        printf("%d\n",ans);    }}