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

时间:2015-11-04 23:08:31      阅读:382      评论:0      收藏:0      [点我收藏+]

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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 n1 edges. You want to complete this tree by adding n1 edges. There must be exactly one path between any two nodes after adding. As you know, there are nn2 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 n1 integers f(1),f(2),,f(n1).

1T2015
2n2015
0f(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
 
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
技术分享
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <ctime>
 6 #include <iostream>
 7 #include <map>
 8 #include <set>
 9 #include <algorithm>
10 #include <vector>
11 #include <deque>
12 #include <queue>
13 #include <stack>
14 using namespace std;
15 typedef long long LL;
16 typedef double DB;
17 #define MIT (2147483647)
18 #define MLL (1000000000000000001LL)
19 #define INF (1000000001)
20 #define For(i, s, t) for(int i = (s); i <= (t); i ++)
21 #define Ford(i, s, t) for(int i = (s); i >= (t); i --)
22 #define Rep(i, n) for(int i = (0); i < (n); i ++)
23 #define Repn(i, n) for(int i = (n)-1; i >= (0); i --)
24 #define mk make_pair
25 #define ft first
26 #define sd second
27 #define puf push_front
28 #define pub push_back
29 #define pof pop_front
30 #define pob pop_back
31 #define sz(x) ((int) (x).size())
32 #define clr(x, y) (memset(x, y, sizeof(x)))
33 inline void SetIO(string Name)
34 {
35     string Input = Name + ".in";
36     string Output = Name + ".out";
37     freopen(Input.c_str(), "r", stdin);
38     freopen(Output.c_str(), "w", stdout);
39 }
40 
41 inline int Getint()
42 {
43     char ch =  ;
44     int Ret = 0;
45     bool Flag = 0;
46     while(!(ch >= 0 && ch <= 9))
47     {
48         if(ch == -) Flag ^= 1;
49         ch = getchar();
50     }
51     while(ch >= 0 && ch <= 9)
52     {
53         Ret = Ret * 10 + ch - 0;
54         ch = getchar();
55     }
56     return Ret;
57 }
58 
59 const int N = 2050;
60 int n, F[N];
61 int Dp[N];
62 
63 inline void Solve();
64 
65 inline void Input()
66 {
67     int TestNumber = Getint();
68     while(TestNumber--)
69     {
70         n = Getint();
71         For(i, 1, n - 1) F[i] = Getint();
72         Solve();
73     }
74 }
75 
76 inline void Solve()
77 {
78     For(i, 2, n - 1) F[i] -= F[1];
79     Dp[0] = 0;
80     For(i, 1, n - 2)
81     {
82         Dp[i] = F[i + 1];
83         For(j, 1, (i / 2) + 1)
84             Dp[i] = max(Dp[i], Dp[j] + Dp[i - j]);
85     }
86     printf("%d\n", Dp[n - 2] + n * F[1]);
87 }
88 
89 int main()
90 {
91     Input();
92     //Solve();
93     return 0;
94 }
View Code

 

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

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原文地址:http://www.cnblogs.com/StupidBoy/p/4937416.html

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