2015ACM/ICPC亚洲区长春站 H hdu 5534 Partial Tree

Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 228    Accepted Submission(s): 138

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）

 

题意：一个树的权值被定义为sigma(f(di), 1<=i<=n)其中di是第i个点的度数，求一个n个点的数的最大权值

分析：先把每个点的度数当成1，然后问题转化为总度数为2n-2，现有度数n，求增加n-2度数的最大权值

dp[i]表示增加i度数的最大权值

dp[i] = max(f[i+1], dp[j]+dp[i-j]), 1<=j<=i/2

完了

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <ctime>
 #include <iostream>
 #include <map>
 #include <set>
 #include <algorithm>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define MIT (2147483647)
 #define MLL (1000000000000000001LL)
 #define INF (1000000001)
 #define For(i, s, t) for(int i = (s); i <= (t); i ++)
 #define Ford(i, s, t) for(int i = (s); i >= (t); i --)
 #define Rep(i, n) for(int i = (0); i < (n); i ++)
 #define Repn(i, n) for(int i = (n)-1; i >= (0); i --)
 #define mk make_pair
 #define ft first
 #define sd second
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define sz(x) ((int) (x).size())
 #define clr(x, y) (memset(x, y, sizeof(x)))
 inline void SetIO(string Name)
 {
     string Input = Name + ".in";
     string Output = Name + ".out";
     freopen(Input.c_str(), "r", stdin);
     freopen(Output.c_str(), "w", stdout);
 }

 inline int Getint()
 {
     char ch = ' ';
     int Ret = ;
     bool Flag = ;
     while(!(ch >= '' && ch <= ''))
     {
         if(ch == '-') Flag ^= ;
         ch = getchar();
     }
     while(ch >= '' && ch <= '')
     {
         Ret = Ret *  + ch - '';
         ch = getchar();
     }
     return Ret;
 }

 const int N = ;
 int n, F[N];
 int Dp[N];

 inline void Solve();

 inline void Input()
 {
     int TestNumber = Getint();
     while(TestNumber--)
     {
         n = Getint();
         For(i, , n - ) F[i] = Getint();
         Solve();
     }
 }

 inline void Solve()
 {
     For(i, , n - ) F[i] -= F[];
     Dp[] = ;
     For(i, , n - )
     {
         Dp[i] = F[i + ];
         For(j, , (i / ) + )
             Dp[i] = max(Dp[i], Dp[j] + Dp[i - j]);
     }
     printf("%d\n", Dp[n - ] + n * F[]);
 }

 int main()
 {
     Input();
     //Solve();
     return ;
 }