Root the Tree Solution

Problem

A directed tree is a directed graph such that if all edges were undirected, this graph would be a tree. A rooted directed tree is a directed tree in which there is one vertex (the root, let's denote it by $r$) such that it is possible to reach all vertices of the graph from $r$ by moving along the directed edges.

You are given a directed tree with $N$ vertices (numbered $1$ through $N$). You may perform the following operation on it any number of times (including zero):

• Choose some edge which currently exists in the graph.
• Remove this edge.
• Add a new edge in such a way that the graph becomes a directed tree again.

What is the smallest number of operations which need to be performed in order to make the directed tree rooted?

Input

• The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
• The first line of each test case contains a single integer $N$.
• Each of the next $N-1$ lines contains two space-separated integers $u$ and $v$ denoting that there is a directed edge from $u$ to $v$ in the tree.

Output

For each test case, print a single line containing one integer ― the smallest number of operations we need to perform to create a rooted directed tree.

Constraints

• $1 \le N \le 10^4$
• $1 \le u, v \le N$
• the graph described on the input is a directed tree
• the sum of $N$ over all test cases does not exceed $10^5$

Subtasks

Subtask #1 (20 points): for each edge, $u = 1$ or $v = 1$ (the graph is a directed star)

Subtask #2 (80 points): original constraints

Sample 1:

Input

Output

2
3
1 2
3 2
3
1 2
2 3

1
0

Explanation:

Example case 1: We can delete the edge from vertex $3$ to vertex $2$ and insert an edge from $3$ to $1$. Then, the graph becomes a rooted directed tree with vertex $3$ as the root. However, there are many other possible solutions.

Example case 2: The graph is already a rooted directed tree; the root is vertex $1$.

 

Program :

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

int comp(const void *p,const void *q)

{ 

    if(*(long int *)p-*(long int *)q>0)

        return 1;

    else 

        return -1;

} 

int main(void) {

// your code goes here

int t,u,v,n,arr[10000],i,ans;

scanf("%d",&t);

while(t--)

{ 

   scanf("%d",&n);

  for(i=0;i<=n;i++) 

    arr[i]=0;

  for(i=1;i<n;i++)

  { 

      scanf("%d %d",&u,&v);

      arr[v]++;

  }

  ans=-1;

  for(i=1;i<=n;i++)

  if(arr[i]==0) 

  ans++;

  printf("%d\n",ans); 

}

return 0;

}