树状dp Tree of Tree

最新推荐文章于 2025-03-04 17:06:05 发布

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本文介绍了一种算法，用于解决给定带权节点树中寻找指定大小的最大子树的问题。通过深度优先搜索（DFS）策略，递归地计算以每个节点为根的最大子树的总权重。

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

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
using namespace std;
const int MAXN=105;
vector<int> tree[MAXN];  
int n,k;
int dp[MAXN][MAXN];  //dp[i][k]表示以i为根节点的最大的k个节点的和 
int vis[MAXN];
int  maxn;
void dfs(int u)
{
	vis[u]=1;
	for(int i=0;i<tree[u].size();i++)
	{
		int v=tree[u][i];
		if(!vis[v])
		{
			dfs(v);
			for(int j=k;j>=1;j--)  //每个根节点不仅要找出k个节点和的最大值，还要算出1到k-1的 对应的节点和的最大值，是为了给后面的根节点用 
			for(int m=1;m<j;m++)
			dp[u][j]=max(dp[u][j],dp[u][m]+dp[v][j-m]); //根节点的m个和 加上 儿子的j-m个和 一共j个和  不断地更新dp[u][j] 
		}
	}
}
int main()
{
  
    while(scanf("%d%d",&n,&k)!=EOF)
    {
    	maxn=0;
    	for(int i=0;i<n;i++)    //每次一定不要忘记将树清空！！！ 
    	tree[i].clear();
	    memset(dp,0,sizeof(dp));
	    memset(vis,0,sizeof(vis));
	    for(int i=0;i<n;i++)
	    	scanf("%d",&dp[i][1]);
	    for(int i=0;i<n-1;i++)
	    {
	        int u,v;
	        scanf("%d%d",&u,&v);
	        tree[u].push_back(v);
	        tree[v].push_back(u);
	    }
	    dfs(0);
	    for(int i=0;i<n;i++)
	    {
	    	if(dp[i][k]>maxn) maxn=dp[i][k];
	    }
	    printf("%d\n",maxn);

    }
   
    return 0;
}