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ＰＯＪ２４８６－－－Apple Tree

2024-10-21 06:42:39   213人阅读

Apple Tree
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7598 Accepted: 2548

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

Source


树形ｄｐ

设dp[i][j][0] 表示　在以ｉ为根的子树下走了ｊ步最后不回到ｉ，能够得到的最多苹果数

dp[i][j][1] 表示　在以ｉ为根的子树下走了ｊ步最后回到ｉ，能够得到的最多苹果数


设ｕ为某子树根，ｖ为其中一个儿子节点

dp[u][i][1] = max(dp[u][i][1], dp[v][j - 2][1] + dp[u][i - j][1] + apple[v]);

dp[u][i][0] = max(dp[u][i][0], dp[u][i - j][1] + dp[v][j - 1][0] + apple[v]);

dp[u][i][0] = max(dp[u][i][0], dp[u][i - j][0] + dp[v][j - 2][1] + apple[v]);


/*************************************************************************
    > File Name: POJ2486.cpp
    > Author: ALex
    > Mail: 405045132@qq.com 
    > Created Time: 2015年01月02日 星期五 11时30分01秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

vector <int> edge[110];
int dp[110][220][2];
int apple[110];
int n, m;

void dfs(int u, int fa)
{
	int size = edge[u].size();
	for (int i = 0; i < size; ++i)
	{
		int v = edge[u][i];
		if (v == fa)
		{
			continue;
		}
		dfs(v, u);
		for (int j = m; j >= 1; --j)
		{
			for (int k = 1; k <= j; ++k)
			{
				if (k >= 2)
				{
					dp[u][j][1] = max(dp[u][j][1], dp[u][j - k][1] + dp[v][k - 2][1] + apple[v]);
					dp[u][j][0] = max(dp[u][j][0], dp[v][k - 2][1] + dp[u][j - k][0] + apple[v]);
				}
				dp[u][j][0] = max(dp[u][j][0], dp[u][j - k][1] + dp[v][k - 1][0] + apple[v]);
			}
		}
	}
}

int main()
{
	int u, v;
	while (~scanf("%d%d", &n, &m))
	{
		memset (dp, 0, sizeof(dp));
		for (int i = 1; i <= n; ++i)
		{
			edge[i].clear();
			scanf("%d", &apple[i]);
		}
		for (int i = 1; i <= n - 1; ++i)
		{
			scanf("%d%d", &u, &v);
			edge[u].push_back(v);
			edge[v].push_back(u);
		}
		dfs(1, -1);
		printf("%d\n", dp[1][m][0] + apple[1]);
	}
	return 0;
}


ＰＯＪ２４８６－－－Apple Tree


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