1976.Number_of_Ways_to_Arrive_at_Destination.cpp

/*
    written by Pankaj Kumar.
    country:-INDIA
    Institute: National Institute of Technology, Uttarakhand
*/
typedef long long ll ;
typedef unsigned long long ull;
typedef vector<int> vl;
typedef vector<vector<int>> vvl;
#define speed cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);
/*  Abbrevations  */
#define ff first
#define ss second
#define mp make_pair
#define pb push_back
// loops
#define forin(arr,n) for(ll i=0;i<n;i++) cin>>arr[i];
// Some print
#define no cout<<"NO"<<endl;
#define yes cout<<"YES"<<endl;
// sort
#define all(V) (V).begin(),(V).end()
#define srt(V) sort(all(V))
#define srtGreat(V) sort(all(V),greater<ll>())
// some extra
#define printv(v) for(ll i=0;i<ll(v.size());i++){cout<<v[i]<<" ";} cout<<endl;
#define precision(x) cout<<fixed<<setprecision(x);
#define sz(V) ll(V.size())


/* ascii value
A=65,Z=90,a=97,z=122
*/

/* Some syntax 
//Syntax to create a min heap for priority queue
//priority_queue <int, vector<int>, greater<int>>pq;
*/
/*  --------------------MAIN PROGRAM----------------------------*/
// to run ctrl+b
const ll INF=1e18;
const ll mod1=1e9+7;
const ll mod2=998244353;
// Techniques :
// divide into cases, brute force, pattern finding
// sort, greedy, binary search, two pointer
// transform into graph

// Experience :
// Cp is nothing but only observation and mathematics.


//Add main code here

class Solution {
public:
    int min_distance(vector<long long> distance,vector<bool>check){
        long long mini=LONG_MAX,node=0;
        for(ll i=0;i<distance.size();i++){
            if(check[i]==false&&mini>=distance[i]){
                mini=distance[i];
                node=i;
            }
        }
        return node;
    }
    int countPaths(int n, vector<vector<int>>& roads) {
        vector<long long> distance(n,LONG_MAX);
        vector<bool>check(n,false);
        distance[0]=0;
        vector<long long> ways(n,0);
        ways[0]=1;
        for(int i=0;i<n-1;i++){
            long long node=min_distance(distance,check);
            check[node]=true;
            for(auto x:roads){
                int connected_node=-1,cost=0;
                if(x[0]==node){
                    connected_node=x[1];
                    cost=x[2];
                }
                else if(x[1]==node){
                    connected_node=x[0];
                    cost=x[2];
                }
                if(connected_node!=-1&&check[connected_node]==false&&distance[node]!=LONG_MAX){
                    long long temp=distance[node]+cost;
                    
                    if(distance[connected_node]==temp){
                        ways[connected_node]+=ways[node];
                        ways[connected_node]%=mod1;
                    }
                    else if(distance[connected_node]>temp){
                        ways[connected_node]=ways[node];
                        ways[connected_node]%=mod1;

                    }
                    distance[connected_node]=min((long long)distance[connected_node],temp);
                }
            }
        }
        return ways[n-1];

    }
};



/* -----------------END OF PROGRAM --------------------*/
/*
* stuff you should look before submission
* constraint and time limit
* int overflow
* special test case (n=0||n=1||n=2)
* don't get stuck on one approach if you get wrong answer
*/