Prime Path (SPOJ) using bfs.cpp

/*
SPOJ Question link : https://www.spoj.com/problems/PPATH/

The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it's not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.

Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on...

Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
    1033
    1733     
    3733     
    3739     
    3779
    8779
    8179 
    
The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).

Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.

Input:
3
1033 8179
1373 8017
1033 1033

Output:
6
7
0
*/

#include <bits/stdc++.h>
using namespace std;
vector<bool> primes(1e4,true);
vector<vector<int>> adj(1e4);
vector<int> pno;
vector<bool> visited(1e4);
int mindist = -1;

void bfs(int a,int b){
    queue<int> q;
    q.push(a);
    mindist++;
    if(a==b) return;
    while(!q.empty()){
        int n=q.size();
        mindist++;
        while(n--){
            int u = q.front();
            q.pop();
            for(int i:adj[u]){
                if(!visited[i]){
                    if(i==b) return;
                    else q.push(i);
                }
            }
        }
        
        
    }
}
  
void primeGenerate(){
    for(int i=2;i<=100;i++){
        for(int j=i*i;j<=10000;j+=i){
            primes[j]=false;
        }
    }
}

bool isValid(int a,int b){
    int cnt=0;
    while(a>0){
        int c = a%10;
        int d = b%10;
        if(c!=d) cnt++;
        a/=10,b/=10;
    }
    return cnt==1;
}

void buildGraph(){
    for(int i=1000;i<=9999;i++) if(primes[i]) pno.push_back(i);
    for(int i=0;i<pno.size();i++){
        for(int j=i+1;j<pno.size();j++){
            if(isValid(pno[i],pno[j])){
                adj[pno[i]].push_back(pno[j]);
                adj[pno[j]].push_back(pno[i]);
            }
        }
    }
}

int main() {
	primeGenerate();
	buildGraph();
	int t;
	cin>>t;
	while(t--){
	    int a,b;
	    cin>>a>>b;
	    for(int i=1000;i<=9999;i++) visited[i]=false;
	    mindist = -1;
	    bfs(a,b);
	    std::cout << mindist << std::endl;
	}
	return 0;
}