HDU2717 Catch That Cow 【广搜】

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7147    Accepted Submission(s): 2254

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 <stdio.h>
#include <queue>
#include <string.h>
#define maxn 100002
using std::queue;
 
struct Node{
	int pos, step;
};
bool vis[maxn];
 
void move(Node& tmp, int i)
{
	if(i == 0) --tmp.pos;
	else if(i == 1) ++tmp.pos;
	else tmp.pos <<= 1;
}
 
bool check(int pos)
{
	return pos >= 0 && pos < maxn && !vis[pos];
}
 
int BFS(int n, int m)
{
	if(n == m) return 0;
	memset(vis, 0, sizeof(vis));
	queue<Node> Q;
	Node now, tmp;
	now.pos = n; now.step = 0;
	Q.push(now);
	vis[n] = 1;
	while(!Q.empty()){
		now = Q.front(); Q.pop();
		for(int i = 0; i < 3; ++i){
			tmp = now;
			move(tmp, i);
			if(check(tmp.pos)){
				++tmp.step;
				if(tmp.pos == m) return tmp.step;
				vis[tmp.pos] = 1;
				Q.push(tmp);
			}
		}
	}
}
 
int main()
{
	int n, m;
	while(scanf("%d%d", &n, &m) == 2){
		printf("%d\n", BFS(n, m));
	}
	return 0;
}