# Timus - 1018 - Binary Apple Tree - Dynamic Programming

Today, I saw this interesting problem.

Timus - 1018 - Binary Apple Tree

When I started working on this, I didnt know why do we need Dynamic Programming to solve this. My idea was pretty simple.
1. While reading Inputs find the total number of apples
2. Find all the leaf Branches (I mean branches without any branches on them)
3. Get the branch with the least number of apples and remove that
4. Goto Step 2 until N - Q - 1 times. (N is the total number of enumerated points on the tree, Q is the number of branches to keep, Always there will be N - 1 branches on the binary tree. So N - Q - 1 will give the number of branches to be removed)
It was pretty straight forward to implement.
`# include <cstdio># include <map># include <set># include <bitset>using namespace std;int N, Q;unsigned long long Total = 0;struct Branch{ bool operator< (const Branch & branch) const {  if (this->Apples == branch.Apples)  {   if (this->Start == branch.Start)    return this->End < branch.End;   return this->Start< branch.Start;  }  return this->Apples < branch.Apples; } Branch (int a, int b, int c){Start = a; End = b; Apples = c;} int Start, End, Apples;};set <Branch> Edges;bitset <101> Visited;map <int, map <int, int> > Tree;void FindEdges (int current, int parent){ //printf ("%d\t%d\n", current, parent); if (Visited[current]) return; //printf ("%d\t%d Not Visited\n", current, parent); Visited[current] = true; map<int, int>::iterator it = Tree[current].begin(); if (Tree[current].size() == 1)  Edges.insert (Branch (parent, current, Tree[parent][current])); else  for (; it != Tree[current].end(); it++)   if (!Visited[it->first])    FindEdges (it->first, current);}int main(){ //freopen ("Input.txt", "r", stdin); //freopen ("Scratch.txt", "w", stdout); scanf ("%d%d", &N, &Q); Q = N - 1 - Q; int A, B, C; for (int i = 0; i < N - 1; i++) {  scanf ("%d%d%d", &A, &B, &C);  Tree[A][B] = C;  Tree[B][A] = C;  Total += C; } for (int i = 0; i < Q; i++) {  Edges.clear();  Visited.reset();  map<int, int>::iterator it1 = Tree.begin();  Visited = true;  for (; it1 != Tree.end(); it1++)   FindEdges(it1->first, 1);  //printf ("Edges Size : %lu\n", Edges.size());  set<Branch>::iterator it = Edges.begin();  //printf ("%d\t%d\t%d\n", it->Start, it->End, it->Apples);  Total -= it->Apples;  Tree[it->Start].erase (it->End);  Tree[it->End].erase (it->Start); } printf ("%llu\n", Total);}`
However, this solution was failing so many times. Then I found the testcase for which this program was failing.
`4 11 2 11 3 22 4 3`
The above shown solution returns 1 as the result whereas the optimal result is 2. This approach is greedy approach I believe.
In the first iteration, it detects the branches
`1 3 22 4 3`
and removes the branch 1 3 2, since it has the least number of apples and in the next iteration it detects the following branch
`2 4 3`
since that is the only leaf branch available and it removes that, leaving us with the suboptimal solution 1.

Here is the DP solution which can really solve this problem and this got accepted by Timus.

`#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;int N,Q, Tree, DP;int solve (int current, int parent, int q){ if (q <= 0) return 0; int lindex = -1, rindex = -1; int & result = DP[current][q]; if (result != -1) return result; for (int i = 0; i < 101; i++)  if (Tree[current][i] != -1 && i != parent) {lindex = i; break;} for (int i = (lindex == -1?0:lindex+1); i < 101; i++)  if (Tree[current][i] != -1 && i != parent) {rindex = i; break;} //printf ("%d\t%d\t%d\t%d\t%d\n", current, parent, lindex, rindex, q); if (lindex == -1 || rindex == -1) return 0; for (int i = 0; i <= q; i++)  result = max (result, (i == q?0:Tree[current][lindex] + solve(lindex, current, q - i - 1))   + (i == 0?0:Tree[current][rindex] + solve(rindex, current, i - 1))); //printf ("Returning %d\n", result); return result;}int main(){ //freopen ("Input.txt", "r", stdin); //freopen ("Scratch.txt", "w", stdout); scanf("%d%d",&N,&Q); memset (Tree, -1, sizeof Tree); memset (DP, -1, sizeof DP); for (int i = 0; i < N; i++)  for (int j = 0, x, y, z; j < N; j++)  {   scanf ("%d%d%d", &x, &y, &z);   Tree [x][y] = z;   Tree [y][x] = z;  } printf ("%d\n", solve (1, 0, Q)); return 0;}`
Please write to me if you want me to explain this program. We can do that over chat.