hdu 5534 Partial Tree 背包DP
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参考:http://www.cnblogs.com/qscqesze/p/4967071.html和http://blog.csdn.net/qq_21057881/article/details/52598441
Partial Tree
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=5534
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
HINT
题意
给你n个点,让你构造出一棵树
假设这棵树最后度数为k的点有num[k]个,那么这棵树的价值为sigma(num[i]*f[i])
其中f[i]是已经给定的
思路:一棵树有2(n-1)个度,每个度都有它的权值,那么就相当于一个容量为2*(n-1)的背包,物品的体积是度数,可是这样有可能会出现没有被选的度数,那么我们就先每个点都分配一个度,然后就是完全背包啦
#include<bits/stdc++.h> using namespace std; const int maxn = 25000; int a[maxn]; int dp[maxn]; int main() { int T,n; scanf("%d",&T); while(T--) { scanf("%d",&n); for(int i = 0;i<n-1;i++) scanf("%d",&a[i]); int V = 2*(n-1)-n; for(int i = 0;i<=n;i++) dp[i]=-1e9; dp[0]=a[0]*n; for(int i=1;i<n-1;i++) a[i]-=a[0]; for(int i = 1;i<=V;i++) for(int j = i;j<=V;j++) dp[j]=max(dp[j],dp[j-i]+a[i]); printf("%d\n",dp[V]); } return 0; }
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