杭电Problem-2717

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10820    Accepted Submission(s): 3374

Problem Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

 

Input

Line 1: Two space-separated integers: N and K

 

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

 

Sample Input

5 17

 

Sample Output

4

Hint
The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.
 

 

#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
struct node {
	int x,step;
}temp, p;
bool vis[10000005];
int d[2] = {1,-1};
int bfs(int a, int b)
{
	memset(vis,false,sizeof(vis));
	queue<node> q;
	while (!q.empty())   q.pop();
	p.x = a;
	p.step = 0;
	q.push(p);
	vis[a] = true;
	while (q.size()) {
		p = q.front();
		q.pop();
		if(p.x == b)  return p.step;
		for(int i = 0; i < 3; i++) {
			if (i == 2)	 temp.x = p.x * 2;
			else	temp.x = p.x + d[i];
			if (temp.x < 0 || vis[temp.x] || temp.x>100000)	continue;
			vis[temp.x] = true;
			temp.step = p.step + 1;
			q.push(temp);
		}
	}
	return -1;
}
int main()
{
	int n, k;
	while (scanf("%d %d", &n, &k) != EOF)
	{
		int ans = bfs(n,k);
		printf("%d
", ans);
	}
	return 0;
}