Closing the Farm 2016 US Open Silver

Problems and Contests
#1

In the problem Closing the Farm (USACO 2016 US Open Silver), my code only runs for the first testcase.

Problem link:
http://www.usaco.org/index.php?page=viewproblem2&cpid=644

USACO Guide link:

My approach was to perform a depth-first search on the graph of the farm to check if it is fully connected for each closed barn. After each DFS, I would then traverse through my visited vector (containing the status of each barn, either visited or not visited). If all the elements in visited are 1 (true), then the current graph is fully connected.

I am now a bit stuck on where the mistake is in my solution/code. Could someone please tell me what I did wrong? Any help would be highly appreciated.

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

int N, M;

vector<int> adj[3001]; //use an adjacency list to store the graph
vector<int> visited; //to keep whether a node is visited or not
vector<int> order; //to keep the order in which the barns will close


//DFS search
void dfs(int s)
{
    if(visited[s])
        return;
   
    visited[s] = 1;
   
    for(auto u: adj[s])
        dfs(u);
}

int main(int argc, char** argv) {
   
    ifstream fin("closing.in");
    ofstream fout("closing.out");
   
    int i, j;
   
    int temp1, temp2;
   
    bool b;
   
    fin >> N >> M;
   
    for(i = 0; i < M; i++) {
        fin >> temp1 >> temp2;
       
        adj[temp1].push_back(temp2);
        adj[temp2].push_back(temp1);
    }
   
    for(i = 0; i < N; i++) {
        fin >> temp1;
       
        order.push_back(temp1);
    }
  	
		//check if the graph is fully connected before any closed barns
    visited = vector<int>(3001);
    dfs(1);
   
    b = false;
   
    for(i = 1; i <= N; i++) {
        if(visited[i] != 1) {
            b = true;
            break;
        }
    }
   
    if(b)
        fout << "NO" << endl;
    else
        fout << "YES" << endl;
   
 		//check if the graph is fully connected for each closed barn
    for(i = 0; i < N - 1; i++) {
        for(j = 1; j <= N; j++)
            visited[j] = 0;
       
        visited[order[i]] = 1;
       
        dfs(order[N-1]);
       
        b = false;
       
        for(j = 1; j <= N; j++)
            if(visited[j] != 1) {
                b = true;
                break;
            }
       
        if(b)
            fout << "NO" << endl;
        else
            fout << "YES" << endl;
    }
    
    return 0;
}
#2