Tree Pruning


Problem Statement :


A tree, t, has n vertices numbered from 1 to n and is rooted at vertex 1. Each vertex i has an integer weight, wi, associated with it, and t's total weight is the sum of the weights of its nodes. A single remove operation removes the subtree rooted at some arbitrary vertex u from tree t.

Given t, perform up to k remove operations so that the total weight of the remaining vertices in t is maximal. Then print t's maximal total weight on a new line.

Note: If t's total weight is already maximal, you may opt to remove 0 nodes.

Input Format

The first line contains two space-separated integers, n and k, respectively.
The second line contains n space-separated integers describing the respective weights for each node in the tree, where the ith integer is the weight of the ith vertex.
Each of the n-1 subsequent lines contains a pair of space-separated integers, u and v, describing an edge connecting vertex u to vertex v.

Constraints
2 <= n <= 10^5
1 <= k <= 200
1 <= i <=n
-10^9 <= wi <= 10^9

Output Format

Print a single integer denoting the largest total weight of t's remaining vertices.



Solution :



title-img


                            Solution in C :

In C++ :





#include <fstream>
#include <iostream>
#include <vector>
#include <bitset>
using namespace std;
const int NMAX = 100004;
const long long INF = 1LL<<60;
vector <int> Tree[NMAX], Level[NMAX];
long long dp[NMAX][201], sum[NMAX];
int n, Father[NMAX], v[NMAX], val[NMAX], First[NMAX], Last[NMAX], ind;
inline void DFS(const int node,const int father){
    First[node] = ++ind;
    v[ind] = node;
    for(vector < int >::iterator it = Tree[node].begin();it != Tree[node].end();++it)
        if(*it != father)
            DFS(*it,node);
    Last[node] = ind;
}

int main(){
    int n, k;
    cin.sync_with_stdio(false);
    cin >> n >> k;
    for(int i = 1;i <= n; ++i)
        cin >> val[i];    
    for(int i=1;i<n;++i){
        int x,y;
        cin >> x >> y;
        Tree[x].push_back(y);
        Tree[y].push_back(x);
    }
    DFS(1,0);
    for(int i = 1;i <= n; ++i){
        for(int j=0;j<=k;++j)
            dp[i][j] = -INF;
    }
    dp[1][0] = 0;
    for(int i = 1;i <= n; ++i)
    {
        int node = v[i];
        for(int j = 0;j <= k; ++j)
            if(dp[i][j]!=-INF)
            {
                dp[i + 1][j] =max(dp[i+1][j],dp[i][j]+val[node]);
                if(j+1<=k)
                    dp[Last[node]+1][j+1] =  max(dp[i][j],dp[Last[node]+1][j+1]);
            }
    }
    long long sol = 0;
    for(int j = 0;j <= k;++j)
        sol = max(sol,dp[n+1][j]);
    cout<<sol<<"\n";
    return 0;
}









In Java :





import java.util.List;
import java.io.IOException;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.ArrayList;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Solution {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TreePruning solver = new TreePruning();
		solver.solve(1, in, out);
		out.close();
	}
}

class TreePruning {

	int[] w;
	List<Integer>[] t;
	List<Integer>[] dt;
	int[] cnt;
	long[][] dp;
	int k;

	public void solve(int testNumber, InputReader in, PrintWriter out) {
		int n = in.nextInt();
		k = in.nextInt();
		dp = new long[n][];
		w = new int[n];
		t = new List[n];
		dt = new List[n];
		cnt = new int[n];
		for (int i = 0; i < n; i++) {
			w[i] = in.nextInt();
			t[i] = new ArrayList<>();
			dt[i] = new ArrayList<>();
		}
		for (int i = 0; i < n - 1; i++) {
			int u = in.nextInt() - 1;
			int v = in.nextInt() - 1;
			t[u].add(v);
			t[v].add(u);
		}

		prepare(0, -1);
		dfs(0);

		long res = Long.MIN_VALUE;
		for (int i = 0; i < dp[0].length; i++) {
			res = Math.max(res, dp[0][i]);
		}

		out.println(res);
	}

	void dfs(int u) {
		for (int v : dt[u]) {
			dfs(v);
		}
		long[] d = new long[Math.min(cnt[u], k) + 1];
		for (int v : dt[u]) {
			long[] nd = new long[d.length];
			Arrays.fill(nd, Long.MIN_VALUE / 2);
			for (int i = 0; i < d.length; i++) {
				for (int j = 0; j < dp[v].length && i + j < nd.length; j++) {
					nd[i + j] = Math.max(nd[i + j], d[i] + dp[v][j]);
				}
			}
			d = nd;
		}
		for (int i = 0; i < d.length; i++) {
			d[i] += w[u];
		}
		if (d.length > 0)
			d[1] = Math.max(d[1], 0);
		dp[u] = d;
	}

	void prepare(int u, int p) {
		if (p != -1) {
			dt[p].add(u);
		}
		cnt[u] = 1;
		for (int v : t[u]) {
			if (v != p) {
				prepare(v, u);
				cnt[u] += cnt[v];
			}
		}
	}
}

class InputReader {
	final InputStream is;
	final byte[] buf = new byte[1024];
	int pos;
	int size;

	public InputReader(InputStream is) {
		this.is = is;
	}

	public int nextInt() {
		int c = read();
		while (isWhitespace(c))
			c = read();
		int sign = 1;
		if (c == '-') {
			sign = -1;
			c = read();
		}
		int res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res = res * 10 + c - '0';
			c = read();
		} while (!isWhitespace(c));
		return res * sign;
	}

	int read() {
		if (size == -1)
			throw new InputMismatchException();
		if (pos >= size) {
			pos = 0;
			try {
				size = is.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (size <= 0)
				return -1;
		}
		return buf[pos++] & 255;
	}

	static boolean isWhitespace(int c) {
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	}

}









In C :





#include <stdio.h>
#include <stdlib.h>
typedef struct _node{
  int x;
  struct _node *next;
} node;
void insert_edge(int x,int y);
void dfs(int x);
long long max(long long x,long long y);
int a[100000],b[100000],size[100000],trace[100000]={0},NN=0;
long long dp[100001][201];
node *table[100000]={0};

int main(){
  int N,K,x,y,i,j;
  long long sum;
  scanf("%d%d",&N,&K);
  for(i=0;i<N;i++)
    scanf("%d",a+i);
  for(i=0;i<N-1;i++){
    scanf("%d%d",&x,&y);
    insert_edge(x-1,y-1);
  }
  dfs(0);
  for(i=0;i<=K;i++)
    dp[0][i]=0;
  for(i=1,sum=0;i<=N;i++){
    sum+=b[i-1];
    for(j=0;j<=K;j++)
      dp[i][j]=sum;
  }
  for(i=1,sum=0;i<=N;i++)
    for(j=0;j<=K;j++){
      if(j!=K)
        dp[i+size[i-1]-1][j+1]=max(dp[i+size[i-1]-1][j+1],dp[i-1][j]);
      dp[i][j]=max(dp[i][j],dp[i-1][j]+b[i-1]);
    }
  printf("%lld",dp[N][K]);
  return 0;
}
void insert_edge(int x,int y){
  node *t;
  t=(node*)malloc(sizeof(node));
  t->x=y;
  t->next=table[x];
  table[x]=t;
  t=(node*)malloc(sizeof(node));
  t->x=x;
  t->next=table[y];
  table[y]=t;
  return;
}
void dfs(int x){
  node *t;
  int i=NN;
  trace[x]=1;
  b[NN++]=a[x];
  for(t=table[x];t;t=t->next)
    if(!trace[t->x])
      dfs(t->x);
  size[i]=NN-i;
  return;
}
long long max(long long x,long long y){
  return (x>y)?x:y;
}









In Python3 :





#!/bin/python3

import os
import sys

#
# Complete the treePrunning function below.
#

from collections import defaultdict

INF = -(1e15)

def dfs(x, f, g, k, weights):
    dpc = [INF]*(k+1)
    dpc[0] = weights[x]
    
    for n in g[x]:
        if n == f:
            continue
        dpn = dfs(n, x, g, k, weights)
        dptmp = [INF]*(k+1)
        for i in range(k+1):
            if dpc[i] == INF:
                break
            for j in range(0, k-i+1):
                if dpn[j] == INF:
                    break
                dptmp[i+j] = max(dptmp[i+j], dpc[i]+dpn[j])
            if i+1 <= k:
                dptmp[i+1] = max(dptmp[i+1], dpc[i])
        dpc = dptmp
    return dpc

def treePrunning(k,weights,edges):
    g = defaultdict(list)
    for u, v in edges:
        g[u-1].append(v-1)
        g[v-1].append(u-1)

    dpn = dfs(0, -1, g, k, weights)
    return max(max(dpn),0)


    
if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    nk = input().split()

    n = int(nk[0])

    k = int(nk[1])

    weights = list(map(int, input().rstrip().split()))

    tree = []

    for _ in range(n-1):
        tree.append(list(map(int, input().rstrip().split())))

    result = treePrunning(k, weights, tree)

    fptr.write(str(result) + '\n')

    fptr.close()
                        








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