Binary Apple Tree

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 integers 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: 
2   5
 \ / 
  3   4
   \ /
    1
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 (2 ≤  N ≤ 100; 1 ≤  Q ≤  N − 1).  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



题意：一颗二叉苹果树树上结苹果，要求对其进行剪枝，求保留Q根树枝能保留的最多到苹果数。
树型dp，当年树型dp入门就是用的这个题，
f[i][j]表示以i为根节点的子树保留j条边所能得到的最多的苹果
 f[root][j + k + 1] = max(f[left][j] + f[right][k] + a[root][i]); （left和right是root的儿子） 

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

const int M = 1100;
int n, m;

long f[M][M];
long a[M][M], map[M][M];
void init()
{
    for(long i = 1; i <= n - 1; i++)
    {
        long x, y, z;
        scanf("%d%d%d",&x, &y, &z);
        map[x][0]++;
        map[x][map[x][0]] = y;
        a[x][map[x][0]] = z;
        map[y][0]++;
        map[y][map[y][0]] = x;
        a[y][map[y][0]] = z;
    }
    for (long i = 1; i <= 1000;  i++)
    for (long j = 1; j <= 1000;  j++)
        f[i][j] = -1;
}
long max(long a, long b)
{
    return a > b ? a : b;
}
void dp(long root, long fa)
{
    //cout <<root <<' '<<fa<<endl;
    f[root][0] = 0;
    for (long i = 1; i <= map[root][0];  i++)
    {
        if (map[root][i] != fa)
        {
            long tmp = map[root][i];
            dp(tmp , root);

           for (long j = m; j >= 0; j--)
            {
                if (f[root][j] >= 0)
                {
                    for (long k = m; k >= 0; k--)
                    {
                        if((j + k <= m)&&(f[tmp][k] >= 0))
                        {
                            f[root][j + k + 1] = max(f[root][j + k + 1], f[root][j] + f[tmp][k] + a[root][i]);
                        }
                    }
                }
            }
        }
    }
}
int main()
{
    scanf("%d%d", &n, &m);
    init();
    dp(1, 0);
    printf("%d\n", f[1][m]);
    return 0;
}