A Binary Apple Tree
Time Limit: 1.0 second
Memory Limit: 16 MB
Let's
imagine how apple tree looks in binary computer world. You're right, it
looks just like a binary tree, i.e. any biparous branch splits up to
exactly two new branches. We will enumerate by natural numbers the root
of binary apple tree, points of branching and the ends of twigs. This
way we may distinguish different branches by their ending points. We
will assume that root of tree always is numbered by 1 and all numbers
used for enumerating are numbered in range from 1 to N, where N is the total number of all enumerated points. For instance in the picture below N is equal to 5. Here is an example of an enumerated tree with four branches:
As
you may know it's not convenient to pick an apples from a tree when
there are too much of branches. That's why some of them should be
removed from a tree. But you are interested in removing branches in the
way of minimal loss of apples.
So your are given amounts of apples on a branches and amount of
branches that should be preserved. Your task is to determine how many
apples can remain on a tree after removing of excessive branches.
Input
First line of input contains two numbers: N and Q (1 ≤ Q ≤ N; 1 < N ≤ 100). N denotes the number of enumerated points in a tree. Q denotes amount of branches that should be preserved. Next N−1
lines contains descriptions of branches. Each description consists of a
three integer numbers divided by spaces. The first two of them define
branch by it's ending points. The third number defines the number of
apples on this branch. You may assume that no branch contains more than
30000 apples.
Output
Output should contain the only number — amount of apples that can be preserved. And don't forget to preserve tree's root ;-)
Sample
| input |
output |
5 2
1 3 1
1 4 10
2 3 20
3 5 20
|
21
|
简析:
这是一个简单的树形动态规划问题,大概可以拿来当这类题目的入门训练题.虽然这是URAL上的第一个树形DP,但是我奇怪的是它的通过率并不很高.
对于原题目的图形,用数组value[a][b]表示a,b点间苹果的个数,用nd[p].L,nd[p].R分别表示节点p的左右儿子.通过build_tree(1)获得数组nd[],从而获得整棵树的信息.
接着,用ans[p][i]表示
以节点p为祖宗的子树,保留的枝条不超过i条时所能保留的最多的苹果,状态转移有一下几种情况.
在除去多余枝条的后的图中,
1.
p只与一个儿子相连: ans[p][i]=max(ans[left_son][i-1]+value[left_son][p],ans[right_son][i-1]+value[right_son][p]);
2.
p与两个儿子相连: for (int j=0;j<=i-2;++j)
ans[p][i]=max(ans[p][i],ans[left_son][j]+ans[right_son][i-j-2]+d);
这里,d=value[left_son][p]+value[right_son][p];
算法在o(N*Q*Q)级别
1 #include<iostream>
2 using namespace std;
3 const int MAXN=102;
4 int n,q,value[MAXN][MAXN],ans[MAXN][MAXN];
5 struct node
6 {
7 int l,r;
8 }nd[MAXN];
9
10 void build_tree(int p)
11 {
12 int flg=0;
13 for (int i=1;i<=n;++i)
14 if (value[p][i] && (!nd[i].l))
15 {
16 flg=1;
17 if (nd[p].l==0) nd[p].l=i;
18 else
19 {nd[p].r=i; break;}
20 }
21 if (!flg) return;
22 if (nd[p].l) build_tree(nd[p].l);
23 if (nd[p].r) build_tree(nd[p].r);
24 }
25
26 void calc(int p)
27 {
28 if (!nd[p].l) return;
29 int l=nd[p].l,r=nd[p].r;
30 calc(l);
31 calc(r);
32 ans[p][1]=max(value[l][p],value[r][p]);
33
34 int d=value[l][p]+value[r][p];
35 for (int i=2;i<=q;++i)
36 {
37 ans[p][i]=max(ans[l][i-1]+value[l][p],ans[r][i-1]+value[r][p]);
38 for (int j=0;j<=i-2;++j)
39 ans[p][i]=max(ans[p][i],ans[l][j]+ans[r][i-j-2]+d);
40 }
41 }
42
43
44 int main()
45 {
46 //freopen("data.in","r",stdin);
47 //freopen("data.out","w",stdout);
48 cin >> n >> q;
49 memset(value,0,sizeof(value));
50 for (int i=1;i<n;++i)
51 {
52 int a,b,c;
53 cin >> a >> b >> c;
54 value[a][b]=c;
55 value[b][a]=c;
56 }
57 memset(nd,0,sizeof(nd));
58 build_tree(1);
59 calc(1);
60 cout << ans[1][q] << endl;
61 return 0;
62 }
63