树状dp Tree of Tree

骑猪看日落 2022-06-17 20:19 195阅读 0赞

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;
    }