ural 1018 Binary Apple Tree（树形DP）

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

题目意思：

有一棵苹果树，苹果树的是一棵二叉树，共N个节点，树节点编号为1~N，编号为1的节点为树根，边可理解为树的分枝，每个分支都长着若干个苹果，现在要要求减去若干个分支，保留M个分支，要求这M个分支的苹果数量最多。

输入：

N M

接下来的N-1行是树的边，和该边的苹果数N and M (1 ≤ M < N;1 < N ≤ 100)

输出：

剩余苹果的最大数量。

思路：由于是一颗二叉树，所以可以用树形DP,对于每一点u寻找u的子树，定义dp[u][j]表示在以u为根的子树保留j个分支可以得到的最大苹果数量

最后答案输出dp[1][m]即可

#include <stdio.h>
#include <string.h>
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
int N, Q;
struct node
{
int v;
int w;
} edge[205];
int first[105];
int next[205];
int size;
int dp[105][105];
void add(int u, int v, int w)
{
edge[size].v = v;
edge[size].w = w;
next[size] = first[u];
first[u] = size++;
}
int dfs(int u, int pre)
{
int sum = 0;
for(int i = first[u]; i != -1; i = next[i])
{
int v = edge[i].v;
if(v == pre) continue;
int w = edge[i].w;
sum += dfs(v, u) + 1;
for(int j = min(sum, Q); j >= 1; --j)
{
for(int k = j; k >= 1; --k)
{
dp[u][j] = max(dp[u][j], dp[u][j-k] + dp[v][k-1] + w);
}
}
}
return sum;
}
int main()
{
int u, v, w;
while(scanf("%d%d", &N, &Q) != EOF)
{
memset(first, -1, sizeof(first));
size = 0;
for(int i = 1; i < N; i++)
{
scanf("%d%d%d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
memset(dp, 0, sizeof(dp));
dfs(1, -1);
printf("%d\n", dp[1][Q]);
}
return 0;
}