# POJ 2486 Apple Tree （树形dp）

## Description

Wshxzt is a lovely girl. She likes apple very much. One day HX takes her to an apple tree. There are N nodes in the tree. Each node has an amount of apples. Wshxzt starts her happy trip at one node. She can eat up all the apples in the nodes she reaches. HX is a kind guy. He knows that eating too many can make the lovely girl become fat. So he doesn’t allow Wshxzt to go more than K steps in the tree. It costs one step when she goes from one node to another adjacent node. Wshxzt likes apple very much. So she wants to eat as many as she can. Can you tell how many apples she can eat in at most K steps.

## Input

There are several test cases in the input

Each test case contains three parts.

The first part is two numbers N K, whose meanings we have talked about just now. We denote the nodes by 1 2 … N. Since it is a tree, each node can reach any other in only one route. (1<=N<=100, 0<=K<=200)

The second part contains N integers (All integers are nonnegative and not bigger than 1000). The ith number is the amount of apples in Node i.

The third part contains N-1 line. There are two numbers A,B in each line, meaning that Node A and Node B are adjacent.

Input will be ended by the end of file.

Note: Wshxzt starts at Node 1.

## Output

For each test case, output the maximal numbers of apples Wshxzt can eat at a line.

## Sample Input

2 1
0 11
1 2
3 2
0 1 2
1 2
1 3


## Sample Output

11
2


## 思路

dp[root][k][1] 代表以 root 为根，最多走 k 步并且最终回到 root 所能得到的最大值。

dp[root][k][0] 代表以 root 为根，最多走 k 步并且最终不回到 root 所能得到的最大值。

dp[root][k][0]=max(dp[root][k][0],dp[root][k-s][1]+dp[son][s-1][0]);

dp[root][k][0]=max(dp[root][k][0],dp[root][k-s][0]+dp[son][s-2][1]);

dp[root][k][1]=max(dp[root][k][1],dp[root][k-s][1]+dp[son][s-2][1]);

s 是我们为当前判断的子节点所分配的步数， s-1 是从 rootson 需要一步， s-2 是从 rootson ，然后从 son 返回到 root 需要的两步。

## AC 代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
typedef __int64 LL;

struct node
{
int u,v,next;
} tree[510];

int cnt;
int n,k;
int dp[210][510][2];
int val[210];

void add(int u,int v)
{
tree[cnt].u=u;
tree[cnt].v=v;
}

void dfs(int root,int mark)
{
for(int i=head[root]; i!=-1; i=tree[i].next)
{
int son=tree[i].v;
if(son==mark)continue;
dfs(son,root);
for(int j=k; j>=1; j--)
{
for(int s=1; s<=j; s++)
{
dp[root][j][0]=max(dp[root][j][0],dp[root][j-s][1]+dp[son][s-1][0]);
dp[root][j][0]=max(dp[root][j][0],dp[root][j-s][0]+dp[son][s-2][1]);
dp[root][j][1]=max(dp[root][j][1],dp[root][j-s][1]+dp[son][s-2][1]);
}
}
}
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
memset(dp,0,sizeof(dp));
for(int i=1; i<=n; i++)
{
scanf("%d",val+i);
for(int j=0; j<=k; j++)
dp[i][j][0]=dp[i][j][1]=val[i];
}
cnt=0;
for(int i=1; i<n; i++)
{
int a,b;
scanf("%d%d",&a,&b);