POJ 3278 Catch That Cow

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.

题意很简单，就是农民有三种操作，看最后能否到达牛的位置，直接bfs即可

代码如下：

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
int N, K, dx;
int step = 0;
struct node {
	int N;
	int step;
}pre, nex;
int p[200005];
void bfs() {
	queue<node> q;
	pre.N = N;
	pre.step = 0;
	p[N] = 1;
	q.push(pre);
	while(!q.empty()) {
		nex = q.front();
		q.pop();
		if(nex.N == K) {
			step = nex.step;
			return ;
		}
		dx = nex.N-1;
		if(p[dx] == 0 && dx >= 0 && dx <= 100505) {
			pre.N = dx;
			pre.step = nex.step+1;
			q.push(pre);
			p[pre.N] = 1;
		}
		dx = nex.N+1;
		if(p[dx] == 0 && dx >= 0 && dx <= 100505) {
			pre.N = dx;
			pre.step = nex.step+1;
			q.push(pre);
			p[pre.N] = 1;
		}
		dx = nex.N*2;
		if(p[dx] == 0 && dx >= 0 && dx <= 100505) {
			pre.N = dx;
			pre.step = nex.step+1;
			q.push(pre);
			p[pre.N] = 1;
		}
	}
	return ;
}
int main() {
	
	while(~scanf("%d%d",&N,&K)) {
		memset(p, 0, sizeof(p));
		step = 0;
		bfs();
		printf("%d\n",step);		
	}
	return 0;
}