hdu 5534 Partial Tree

Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 366    Accepted Submission(s): 204

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

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

 首先我们要知道prufer sequence这个东西，那么我们就知道可以用n-2的任意序列来表示n个节点的树，同时我们可以知道序列里决定节点度数的是这个节点在数列里出现了多少次，那么，我们对序列排序之后，结构虽然改变，节点度数不变，于是可以用f[i][j]表示前i个点，最后那个点出现了j次产生的值。f[i][j]=f[i-1][j-1]+v[j+1]-v[j];

f[i][1]=max(f[i-1][j])+v[2]-v[1];f[0][0]=n*v[1];

#include<bits/stdc++.h>using namespace std;const int maxn=2100;const int inf=1000000000;int v[maxn],n,f[maxn][maxn];void work(){    scanf("%d",&n);    for(int i=1;i<n;i++)scanf("%d",v+i);    for(int i=0;i<=n;i++)        for(int j=0;j<=n;j++)f[i][j]=-inf;    f[0][0]=n*v[1];    for(int i=1;i<=n-2;i++){        int tmp=-inf;        for(int j=1;j<=i;j++){            f[i][j]=f[i-1][j-1]+v[j+1]-v[j];            tmp=max(tmp,f[i-1][j-1]);        }        f[i][1]=tmp+v[2]-v[1];    }    int ans=f[n-2][0];    for(int i=1;i<=n;i++)ans=max(ans,f[n-2][i]);    printf("%d\n",ans);}int main(){    int t;    scanf("%d",&t);    while(t--)work();    return 0;}