POJ 2486 Apple Tree

最新推荐文章于 2020-08-19 09:18:00 发布

ShinyaLicone

最新推荐文章于 2020-08-19 09:18:00 发布

阅读量528

点赞数 1

CC 4.0 BY-SA版权

分类专栏： POJ 动规文章标签： C++ 树形DP

本文链接：https://blog.youkuaiyun.com/SenyeLicone/article/details/52690269

动规同时被 2 个专栏收录

160 篇文章

订阅专栏

POJ

105 篇文章

订阅专栏

本文介绍了一种使用树形动态规划解决的问题：在一个有N个节点的树中，每个节点有一定数量的苹果，从特定节点开始，在不超过K步的情况下，如何吃到最多的苹果。文章详细解释了状态定义、状态转移方程，并给出了具体的实现代码。

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

Sample Output

11
2

Source

POJ Contest,Author:magicpig@ZSU

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

树形DP~

因为走到一个叶子节点后要原路返回才能走到下一个节点，所以记录的f数组要多加一维状态记录是否会返回该点，则f[i][j][k]就表示从点i向子节点走了j步，如果k==1，则表示没有回到该点，k==0则表示回到该点。

状态转移方程就是f[u][i][1]=max(f[u][i][1],f[u][i-j][0]+f[kkz][j-1][1]);
f[u][i][1]=max(f[u][i][1],f[u][i-j][1]+f[kkz][j-2][0]);
f[u][i][0]=max(f[u][i][0],f[u][i-j][0]+f[kkz][j-2][0]);

（f数组不能开太大，否则会T的。）

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int n,k,x,y,akz,w[10001],ne[10001],fi[10001],fa[10001],f[101][201][2],cnt;

void add(int u,int v)
{
	w[++cnt]=v;ne[cnt]=fi[u];fi[u]=cnt;
}

void findd(int u)
{
	for(int kkzv=fi[u];kkzv;kkzv=ne[kkzv])
	  if(w[kkzv]!=fa[u])
	  {
		int kkz=w[kkzv];fa[kkz]=u;
		findd(kkz);
		for(int i=k;i>=1;i--)
		  for(int j=1;j<=i;j++)
		  {
		  	f[u][i][1]=max(f[u][i][1],f[u][i-j][0]+f[kkz][j-1][1]);
		  	f[u][i][1]=max(f[u][i][1],f[u][i-j][1]+f[kkz][j-2][0]);
		  	f[u][i][0]=max(f[u][i][0],f[u][i-j][0]+f[kkz][j-2][0]);
		  }
	  }
}

int main()
{
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		memset(f,0,sizeof(f));
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&akz);fi[i]=0;
			for(int j=0;j<=k;j++) f[i][j][0]=f[i][j][1]=akz;
		}
		cnt=0;
		for(int i=1;i<n;i++)
		{
			scanf("%d%d",&x,&y);
			add(x,y);add(y,x);
		}
		findd(1);
		printf("%d\n",max(f[1][k][0],f[1][k][1]));
	}
	return 0;
}