Cut Tree


Problem Statement :


Given a tree T with n nodes, how many subtrees (T') of T have at most K edges connected to (T - T')?

Input Format

The first line contains two integers n and K followed by n-1 lines each containing two integers a & b denoting that there's an edge between a & b.

Constraints

1 <= K <= n <= 50
Every node is indicated by a distinct number from 1 to n.

Output Format

A single integer which denotes the number of possible subtrees.



Solution :



title-img


                            Solution in C :

In C++ :





#include<iostream>
#include<string.h>
using namespace std;

#define MAXN 100
int first[MAXN], next1[MAXN], v[MAXN];
int edgeindex;
int n, k;
long long dp[MAXN][MAXN];
bool vis[MAXN];

void addedge(int a, int b)
{
	v[edgeindex] = b;
	next1[edgeindex] = first[a];
	first[a] = edgeindex++;
}

void solve(int cur, int root)
{
	if(vis[cur])
	    return;
	    
	for(int i = first[cur]; i != -1; i = next1[i])
	{
		if(v[i] == root)
			continue;
		solve(v[i], cur);
		for(int j = k; j > 0; --j)
		{
			int tmp = 0;
			for(int x = j; x > 0; x--)
			{
				dp[cur][j] += dp[cur][j-x] * dp[v[i]][x];
			}
			dp[cur][j] += dp[cur][j-1];
		}
	}
	return;
}

int main()
{
	while(cin>>n>>k)
	{
		edgeindex = 0;
		memset(first, -1, sizeof(first));
		memset(dp, 0, sizeof(dp));
		memset(vis, false, sizeof(vis));
		int a, b;
		for(int i = 1; i < n; ++i)
		{
			cin>>a>>b;
			addedge(a, b);
			addedge(b, a);
		}
		
		for(int i = 1; i <= n; ++i)
		{
			dp[i][0] = 1;
		}
		
		solve(1, -1);
		
		long long ans = 0;
		for(int i = 1; i <= n; ++i)
		{
			for(int j = 0; j < k; ++j)
			{
				ans += dp[i][j];
			}
		}
		ans += dp[1][k];
		cout<<ans+1<<endl;
	}
}








In Java :






import java.io.BufferedInputStream;
import java.util.ArrayList;
import java.util.Scanner;


public class Solution {
	static long rooted[][] = new long [55][55];
	static int size[] = new int[55];
	static int outDegree[] = new int[55];
	static ArrayList<Integer> tree[] = new ArrayList[55];
	static int n;
	static int k;
	
	static void dfs(int u, int p) {
		size[u] = 1;
		
		for (int v : tree[u]) {
			if (v == p) continue;
			dfs(v, u);
			
			outDegree[u]++;
			size[u] += size[v];
		}
		//////////////////////////////////////////////////////////////////
		long pre[] = rooted[u];
		rooted[u][0] = 1;
		for (int v : tree[u]) {
			if (v == p) continue;
		
			rooted[u] = new long[55];
			for (int i=0; i<=k; i++) {
				if (i >= 1) rooted[u][i] += pre[i - 1];
				for (int j=0; j<=i; j++)
					rooted[u][i] += pre[i - j] * rooted[v][j];
			}
			pre = rooted[u];	
		}
	}
	
	public static void main(String[] args) {
		Scanner cin = new Scanner(new BufferedInputStream(System.in));
		n = cin.nextInt();
		k = cin.nextInt();
		for (int i=0; i<n; i++) tree[i] = new ArrayList<Integer>();
		for (int i=0; i<n-1; i++) {
			int u = cin.nextInt(); u--;
			int v = cin.nextInt(); v--;
			tree[u].add(v);
			tree[v].add(u);
		}
		
		dfs(0, -1);

		long ans = 0;
		for (int i=0; i<n; i++)
			if (i == 0)
				for (int j=0; j<=k; j++) ans += rooted[i][j];
			else 
				for (int j=1; j<=k; j++) ans += rooted[i][j - 1];
		System.out.println(ans + 1);
		
//		for (int i=0; i<n; i++) {
//			for (int j=0; j<=k; j++)
//				System.out.print(rooted[i][j] + " ");
//			System.out.println();
//		}
//		System.out.println();
	}
}








In C :





#include <stdio.h>
#include <stdlib.h>
void dfs(int x,int parent);
int table[50][50]={0},c[50]={0},p[50],q[51]={0},N;
long long dp1[50][51]={0},dp2[50][51]={0},temp[51],ans; 
int main(){
  int K,x,y,i,j,k;
  scanf("%d%d",&N,&K);
  for(i=0;i<N-1;i++){
    scanf("%d%d",&x,&y);
    table[x-1][y-1]=-1;
    table[y-1][x-1]=-1;
  }
  dfs(0,0);
  for(i=0;i<N;i++)
    if(!c[i])
      q[++q[0]]=i;
  while(q[0]){
    x=q[q[0]--];
    dp1[x][0]=dp2[x][0]=1;
    for(i=0;i<N;i++)
      if(table[x][i]==1){
        for(j=1;j<=K;j++)
          temp[j]=0;
        for(j=1;j<=K;j++){
          dp1[x][j]+=dp1[i][j]+dp2[i][j-1];
          for(k=0;k<=j;k++)
            if(k==1)
              temp[j]+=(dp2[i][k]+1)*dp2[x][j-k];
            else
              temp[j]+=dp2[i][k]*dp2[x][j-k];
        }
        for(j=1;j<=K;j++)
          dp2[x][j]=temp[j];
      }
    if(x){
      c[p[x]]--;
      if(!c[p[x]])
        q[++q[0]]=p[x];
    }
  }
  for(i=0,ans=0;i<=K;i++)
    ans+=dp1[0][i]+dp2[0][i];
  printf("%lld",ans);
  return 0;
}
void dfs(int x,int parent){
  int i;
  p[x]=parent;
  for(i=0;i<N;i++)
    if(table[x][i] && i!=parent){
      table[x][i]+=2;
      c[x]++;
      dfs(i,x);
    }
  return;
}








In Python3 :





line = input().split()
n = int(line[0])
K = int(line[1])

adjlst = [[1]]
for i in range(n):
    adjlst.append([])
adjlst[1].append(0)

for i in range(n - 1):
    line = input().split()
    x = int(line[0])
    y = int(line[1])
    adjlst[x].append(y)
    adjlst[y].append(x)

def dfs(node, par, adjlst, K):
    conlst = [0] * (K + 1)
    dislst = [0] * (K + 1)
    conlst[0] = 1
    for chld in adjlst[node]:
        if chld != par:
            tcl, tdl = dfs(chld, node, adjlst, K)
            # print(str(tcl))
            # print(str(tdl))
            dislst[0] += tdl[0]
            tcl2 = [0] * (K + 1)
            for i in range(K):
                dislst[i + 1] += tcl[i] + tdl[i + 1]
                tcl2[i + 1] += conlst[i]
            for i in range(K + 1):
                for j in range(K + 1):
                    if i + j <= K:
                        tcl2[i + j] += tcl[i] * conlst[j]
            conlst = list(tcl2)
    return conlst, dislst

conlst, dislst = dfs(1, 0, adjlst, K)
# print(str(conlst))
# print(str(dislst))
ans = sum(conlst) + sum(dislst) + 1
print(str(ans))
                        








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