HDU5534 Partial Tree 【DP】

Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1305    Accepted Submission(s): 654


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）

---

不难发现 ， 一颗树每个点的度之和 一定 = 2*n-2
可以很容易得到一个$O(n^3)$的dp算法

d[i][j]=前i个点，剩余j个度可以分配，最大的cool值之和是多少
d[i]j]=max(d[i][j-k]+f[k],d[i][j])
d[0][n-2]=0;

考虑怎样降下一维就OK了
不难发现 如果每个点先分配1度
剩余的(n-2)度可以任取，那就变成完全背包问题

d[i]=已经分配了i个度，最大的cool值之和
d[i]=max(d[i],d[i-(j-1)]+f[j]-f[1])
d[0]=n*f[1]

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<string.h>
#include<math.h>
#include<list>

using namespace std;

#define ll long long
#define pii pair<int,int>
const int inf = 1e9 + 7;

inline int read(){
    int x;
    char ch;
    while(!isdigit(ch=getchar()));
    x=ch-'0';
    while(isdigit(ch=getchar())){
        x=x*10+ch-'0';
    }
    return x;
}

const int N = 3e3+5;

int f[N];
int d[N];

int dp(int n){
    fill(d,d+n,-inf);
    d[0]=n*f[1];
    for(int i=2;i<=n-1;++i){
        for(int j=i-1;j<=n-2;++j){
            d[j]=max(d[j],d[j-(i-1)]+f[i]-f[1]);
        }
    }
    return d[n-2];
}

int main()
{
    //freopen("/home/lu/Documents/r.txt","r",stdin);
    //freopen("/home/lu/Documents/w.txt","w",stdout);
    int T=read();
    while(T--){
        int n=read();
        for(int i=1;i<=n-1;++i){
            f[i]=read();
        }
        printf("%d\n",dp(n));
    }
    return 0;
}