hdu5534 Partial Tree  dp(难)

Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 374    Accepted Submission(s): 209

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）

将前n个度数分给n个点，剩下的就可以随便分了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
const int maxn=100100;
const int INF=(1<<29);

int f[maxn],n;
ll dp[maxn];

int main()
{
    int T;cin>>T;
    while(T--){
        cin>>n;
        REP(i,1,n-1) scanf("%d",&f[i]);
        ll ans=f[1]*n;
        MS0(dp);
        REP(i,0,n-2) dp[i]=-INF;
        dp[0]=0;
        REP(i,1,n-2){
            REP(j,0,i){
                dp[i]=max(dp[i],dp[i-j]+f[j+1]-f[1]);
                //cout<<i<<" "<<dp[i]<<endl;
            }
        }
        ans+=dp[n-2];
        printf("%I64d\n",ans);
    }
    return 0;
}