Problem name: Chef and Reversing

Problem link: https://www.codechef.com/problems/REVERSE

Problem statement: Sometimes mysteries happen. Chef found a directed graph with N vertices and M edges in his kitchen! The evening was boring and chef has nothing else to do, so to entertain himself, Chef thought about a question "What is the minimum number of edges he needs to reverse in order to have at least one path from vertex 1 to vertex N, where the vertices are numbered from 1 to N.

Doubt: Greetings everyone, I am trying to solve this problem, I got all the logic and i have coded the problem too, but I have come across a situation in the given code, this code gives a WA, but if I uncomment Line '1' and comment the Line '2', it gives a an AC, can anybody please help me understand the issue

My logic for this is to firstly mark all given edges u->v as true in the map, and then traverse the entire map then I add the every edge u->v with a weight of 0, but if v->u is not present in the map, then I add v->u with a weight of 1

My code:

#include <bits/stdc++.h>
using namespace std;
#define fastio()                      \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define lli long long int
#define ll long long
#define pii pair<int,int>

int INF=1e9+1000;
int N=1e5+100;
vector<vector<pii>> adj(N,vector<pii>());
vector<int> dist(N);
int n,m; 

void add_edge(int u,int v,int w)
{
    adj[u].push_back({v,w});
}

void dijsktra(int src)
{
    for(int i=1;i<=n;i++) dist[i]=INF;
    priority_queue<pii,vector<pii>,greater<pii>> min_heap;
    dist[src]=0;
    min_heap.push({dist[src],src});
    
    while(!min_heap.empty())
    {
        int curr_node=min_heap.top().second;
        int curr_dist=min_heap.top().first;
        min_heap.pop();
        for(auto opt:adj[curr_node])
        {
            int next_node=opt.first;
            int next_dist=opt.second;
            
            if(curr_dist+next_dist<dist[next_node])
            {
                dist[next_node]=curr_dist+next_dist;
                min_heap.push({dist[next_node],next_node});
            }
        }
    }
    
    return;
}

void solve()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++) adj[i].clear();
    
    map<pii,bool> present;
    for(int i=0;i<m;i++)
    {
        int u,v; cin>>u>>v;
        if(u==v) continue;
        present[{u,v}]=true;
        //add_edge(u,v,0); // Line 1  ***********************************
    }
    
    for(auto x:present)
    {
        pii p=x.first;
        add_edge(p.first,p.second,0); // Line 2  ************************
        if(!present[{p.second,p.first}]) add_edge(p.second,p.first,1);
    }
    
    dijsktra(1);
    
    cout<<(dist[n]>=INF ? -1 : dist[n])<<'\n';
    
    return;
}

int main()
{
    fastio();
    int t=1;
    // cin>>t;
    while(t--) solve();

    return 0;
}