BFS 3071. Number Game

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There is an interesting game. We start with a number K. Then three operations is allowed:

1. Multiply current number by 2
2. Divide current number by 2, if current number is even.
3. Increase current number by 1

Our target is to get number P with minimum operations. You can assume 0 < P,K < 100000, and the number can not be greater than or equal to 100000 at any time.

Input

The first line is an integer T, the number of test cases. Then T cases follows.

Each test case contain two numbers K and P in one line separated by a space.

Output

Output the minimum number of operations. If it is impossible to get P from K, output -1 instead.

Sample Input

2
11 20
99999 50000

Sample Output

7
-1
#include<iostream>
using namespace std;
const int n=100000;
int visit[n];
struct num{
	int x,step;
}c[n];
int bfs(int s,int t){
	memset(visit,0,sizeof(visit));
	visit[s]=1;
	c[0].x=s;
	c[0].step=0;
	int top=1,temp;
	for(int i=0;i<top;i++){
		if(c[i].x==t)
			return c[i].step;
		temp=c[i].x*2;
		if(temp<n&&!visit[temp]){
			visit[temp]=1;
			c[top].step=c[i].step+1;
			c[top].x=temp;
			top++;
		}
		if(c[i].x%2==0){
			temp=c[i].x/2;
			if(temp>0&&!visit[temp]){
				visit[temp]=1;
				c[top].step=c[i].step+1;
				c[top].x=temp;
				top++;
			}
		}
		temp=c[i].x+1;
		if(temp<n&&!visit[temp]){
			visit[temp]=1;
			c[top].step=c[i].step+1;
			c[top].x=temp;
			top++;
		}
	}
	return -1;
}
int main(){
	int N,s,t;
	cin>>N;
	while(N--){
		cin>>s>>t;
		cout<<bfs(s,t)<<endl;
	}
	return 0;
}