URAL 1018 Binary Apple Tree(树形dp入门题)

URAL上简单的树形dp，但是都错了题意。。。sad，是保留m个树干，不是删掉m个树干。求保留的最多的苹果有多少。根据树的特点，每个dp[i][j]表示第i个节点保留了j个树干。dp[i][j] = max(dp[i][j], dp[x1][k-1]+map[x1][i] + dp[x2][j-k-1] + map1[x2][i]).就是多的这个节点是属于哪个子树上的。

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

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
//#define LL __int64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

const int maxn = 110;
using namespace std;

vector<int>g[maxn];
vector<int>now[maxn];
int dp[maxn][maxn];
int vis[maxn];
int map1[maxn][maxn];
int n, m;

void change(int x)
{
vis[x] = 1;
for(int i = 0; i < g[x].size(); i++)
{
if(vis[g[x][i]])
continue;
now[x].push_back(g[x][i]);
change(g[x][i]);
}
}

void dfs(int x)
{
dp[x][0] = 0;
for(int i = 0; i < now[x].size(); i++)
dfs(now[x][i]);
if(now[x].size() == 0)
return;
int ll = now[x][0];
int rr = now[x][1];
for(int i = 1; i <= m; i++)
{
for(int j = 0; j <= i; j++)
{
int xx = 0;
if(j != 0)
xx += (dp[ll][j-1]+map1[x][ll]);
if(i-j != 0)
xx += (dp[rr][i-j-1]+map1[x][rr]);
dp[x][i] = max(dp[x][i],xx);
}
}
}

int main()
{
while(cin >>n>>m)
{
for(int i = 0; i <= n; i++)
{
now[i].clear();
g[i].clear();
map1[i][i] = 0;
}
memset(dp, 0 , sizeof(dp));
memset(vis, 0 , sizeof(vis));
int u, v, w;
for(int i = 0; i < n-1; i++)
{
cin >>u>>v>>w;
g[u].push_back(v);
g[v].push_back(u);
map1[u][v] = w;
map1[v][u] = w;
}
change(1);
dfs(1);
cout<<dp[1][m]<<endl;
}
}