HDU 5534 Partial Tree (变形完全背包 好题)

Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 462    Accepted Submission(s): 236

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?
 
[align=left]Input[/align]

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.
 
[align=left]Output[/align]

For each test case, please output the maximum coolness of the completed tree in one line.
 
[align=left]Sample Input[/align]

2
3
2 1
4
5 1 4

 
[align=left]Sample Output[/align]

5
19

 
[align=left]Source[/align]

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5534

题目大意：构造一棵树，已知度为i的点的值为f[i]，现在求怎么构造这个树可以使总的数值最大，求最大的数值

题目分析：长春银牌题，这题第一反应是二维n3方的背包，显然做不了，一棵n个点的树的总度数为2n-2，并且每个点的度数至少为1，因此我们可以先给每个点一个度为1的值，总值即n*f[1]，然后对剩下的n－2的度做完全背包(dp[i]表示用了i度能得到的最大价值)，再分配的时候每个点已经有了一个度，要变换一下度值关系，而且dp要初始化为负无穷，因为变换后的f可能小于0(比如度为2的点的权值比度为1的小)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 2200;
int const INF = 0x3fffffff;
int dp[MAX], f[MAX], n;

int main()
{
int T;
scanf("%d", &T);
while(T --)
{
int n;
scanf("%d", &n);
for(int i = 0; i <= n; i++)
dp[i] = -INF;
for(int i = 1; i <= n - 1; i++)
scanf("%d", &f[i]);
int val1 = n * f[1];
for(int i = 2; i <= n - 1; i++)
f[i] -= f[1];
for(int i = 1; i <= n - 2; i++)
f[i] = f[i + 1];
dp[0] = 0;
for(int i = 1; i <= n - 2; i++)
for(int j = i; j <= n - 2; j++)
dp[j] = max(dp[j], dp[j - i] + f[i]);
printf("%d\n", dp[n - 2] + val1);
}
}