POJ 2486 Apple Tree

Apple Tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6148		Accepted: 1986

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

题意：有一颗苹果树，每个节点上面有很多苹果，从一个节点到另外一个可以到达的节点花费1步，求k步最多能吃到多少苹果。

这是典型的回溯型树状dp。dp[0][i][j]代表以i为根节点的子树最多j步后回到i能吃到的最多的苹果，dp[1][i][j]代表以i为根节点的子树最多j步后不回到i节点最多能吃到的子树。那么状态转移就分三步了。

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

（2）dp[1][i][j]] = max(dp[1][i][j], dp[0][i][j-k]+dp[1][son-1][k]);  人留在i的子节点son的子树中

（3）dp[1][i][j] = max(dp[1][i][j], dp[1][i][j-k]+dp[0][son-2][k]);  人留在不是son的i的子节点的子树中

#include<cstdio>
#include<cstring>
#include<vector>

using namespace std;

#define Max_(a,b)	(a>b?a:b)

int N,K;
int Val[200],S[200];
bool vis[200];
int d[2][200][300];
vector<int> Son[200];

void DFS(int x)
{
	int i,j,u,m;
	d[0][x][0]=Val[x];
	d[1][x][0]=Val[x];
	for(i=0;i<Son[x].size();i++)
	{
		if(vis[Son[x][i]])
			continue;
		u=Son[x][i];
		vis[u]=1;
		DFS(u);
		for(j=K;j>=0;j--)
		{
			for(m=1;m<=j;m++)
			{
				if(m>=2)
				{
					d[0][x][j]=Max_(d[0][x][j],d[0][x][j-m]+d[0][u][m-2]);
					d[1][x][j]=Max_(d[1][x][j],d[1][x][j-m]+d[0][u][m-2]);
				}
				d[1][x][j]=Max_(d[1][x][j],d[0][x][j-m]+d[1][u][m-1]);
			}
		}
	}
}

int main()
{
	int a,b;
	int i;
	while(scanf("%d%d",&N,&K)==2)
	{
		for(i=1;i<=N;i++)
		{
			scanf("%d",&Val[i]);
			Son[i].clear();
		}
		for(i=1;i<N;i++)
		{
			scanf("%d%d",&a,&b);
			Son[a].push_back(b);
			Son[b].push_back(a);
		}

		memset(vis,0,sizeof(vis));
		memset(d,0,sizeof(d));
		vis[1]=1;
		DFS(1);

		int max=0;
		for(i=0;i<=K;i++)
		{
			max=Max_(max,d[0][1][i]);
			max=Max_(max,d[1][1][i]);
		}
		printf("%d\n",max);
	}

	return 0;
}