1018. Binary Apple Tree

1018. Binary Apple Tree

Time limit: 1.0 second
Memory limit: 64 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 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

Problem Source: Ural State University Internal Contest '99 #2 

这道题目是第一道树形dp，比起一般的dp，感觉有那么一点难度。首先如何存储二叉树，为了方便dp，用l和r数组记录左孩子和右孩子。并且此题比较麻烦，题目给的是边权，不易dp，于是将边权下移到下方节点，但此时根节点就没有了权值，故定义为0；

假设dp[i][j]表示以i为根有j个节点的子树的最大apple数量。

那么有转移方程dp[i][j]=max{dp[l[i]][k]+dp[r[i]][j-1-k]}+v[i] 0<=k<j. 注意dp[][0]=0并且由于叶子节点没有孩子，这里初始化l和r为0，所以dp[0][]=0. 树形dp感觉不好写f递推形式，故用了记忆化搜索。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#define Maxn 110
using namespace std;

int adj[Maxn][Maxn],vis[Maxn],l[Maxn],r[Maxn];
int dp[Maxn][Maxn],v[Maxn];
void dfs(int u,int n){
    vis[u]=1;
    for(int i=1;i<=n;i++){
        if(vis[i]) continue;
        if(adj[u][i]!=-1){
            if(!l[u]) l[u]=i;
            else r[u]=i;
            v[i]=adj[u][i];
            dfs(i,n);
        }
    }
}
int dfs1(int x,int y){
    if(x==0||y==0) return 0;
    int &ans=dp[x][y];
    if(ans) return ans;
    for(int i=0;i<y;i++)
        ans=max(ans,dfs1(l[x],i)+dfs1(r[x],y-1-i)+v[x]);
    return ans;
}
int main()
{
    int n,m,a,b,w;
    scanf("%d%d",&n,&m);
    memset(adj,-1,sizeof adj);
    memset(vis,0,sizeof vis);
    memset(l,0,sizeof l);
    memset(r,0,sizeof r);
    memset(dp,0,sizeof dp);
    v[1]=0;
    for(int i=1;i<n;i++){
        scanf("%d%d%d",&a,&b,&w);
        adj[a][b]=adj[b][a]=w;
    }
    dfs(1,n);
    printf("%d\n",dfs1(1,++m));
    return 0;
}