Tree of Tree ---ZOJ - 3201--树形背包dp

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于 2019-08-17 20:39:21 发布

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本文链接：https://blog.youkuaiyun.com/qq_43625313/article/details/99697512

本文介绍了一种使用树形动态规划解决寻找指定大小的最大子树问题的方法。通过定义状态dp[root][j]，表示节点root的子树中，包含j个节点时的权值总和，利用类似背包的状态转移方程，实现高效求解。

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链接： https://vjudge.net/problem/ZOJ-3201

Describtion

You’re given a tree with weights of each node, you need to find the maximum subtree of specified size of this tree.

Tree Definition
A tree is a connected graph which contains no cycles.

Input

There are several test cases in the input.

The first line of each case are two integers N(1 <= N <= 100), K(1 <= K <= N), where N is the number of nodes of this tree, and K is the subtree’s size, followed by a line with N nonnegative integers, where the k-th integer indicates the weight of k-th node. The following N - 1 lines describe the tree, each line are two integers which means there is an edge between these two nodes. All indices above are zero-base and it is guaranteed that the description of the tree is correct.

Output

One line with a single integer for each case, which is the total weights of the maximum subtree.

Sample Input
3 1
10 20 30
0 1
0 2
3 2
10 20 30
0 1
0 2

Sample Output
30
40

题意：寻找一棵节点数为 k 的子树，保证其节点的权值和最大；

思路：树形dp，不过这里比较特殊，用了类似于背包的写法； dp [ root ] [ j ]表示的是在节点为 root 时，其子树的节点数为 j 个时的权值和；

状态转移方程：
dp[root ] [ j ] = max ( dp[ root ][ j ],dp[ root ][ j-k ]+dp[ child ] [k] );

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
int n,m,dp[110][110];// dp[root][j]表示的是在节点为 root 时，其子树的节点数为 j 个时的权值和； 
vector<int > tree[110];
void dfs(int root,int father)// root当前节点和父亲节点； 
{
	for(int i=0;i<tree[root].size();i++)
	{
		int child=tree[root][i];
		if(child==father) continue; //保证每个节点遍历一次，不重复； 
		dfs(child,root);
		for(int j=m;j>=0;j--) //背包； 从后往前； 保证状态只用一次； 
		{
			for(int k=1;k<j;k++)
			dp[root][j]=max(dp[root][j],dp[root][j-k]+dp[child][k]);
		}
	}
	return ;
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(int i=0;i<=n;i++) 
		tree[i].clear();
		memset(dp,0,sizeof(dp));
		for(int i=0;i<n;i++) scanf("%d",&dp[i][1]);
		for(int i=1;i<=n-1;i++)
		{
			int x,y;
			scanf("%d%d",&x,&y);
			tree[x].push_back(y);
			tree[y].push_back(x);
		}
		dfs(1,-1);
		int ma=-1;
		for(int i=0;i<n;i++) ma=max(ma,dp[i][m]); //寻找最大值； 
		printf("%d\n",ma);
	}
	return 0;
}