POJ 3278 Catch That Cow

Problem

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

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN = 100010;
int n, k;
struct node {//记录当前位置和步数
	int x, step;
};
queue<node> Q;
int vis[MAXN];
void bfs() {
	int x1, s1;
	while(!Q.empty()) {
		node temp = Q.front();
		Q.pop();
		x1 = temp.x;
		s1 = temp.step;
		if(x1 == k) {
			printf("%d\n", s1);
			return;
		}
		if(x1 >= 1 && !vis[x1 - 1]) {//x-1的位置
			node t;
			vis[x1 - 1] = 1;
			t.x = x1 - 1;
			t.step = s1 + 1;
			Q.push(t);
		}
		if(x1 <= k && !vis[x1 + 1]) {
			node t;
			vis[x1 + 1] = 1;
			t.x = x1 + 1;
			t.step = s1 + 1;
			Q.push(t);
		}
		if(x1 <= k && !vis[x1*2]) {
			node t;
			vis[x1*2] = 1;
			t.x = x1*2;
			t.step = s1 + 1;
			Q.push(t);
		}
	}
}
int main() {
	while(scanf("%d %d", &n, &k) != EOF) {
		while(!Q.empty()) Q.pop();
		memset(vis, 0, sizeof(vis));
		vis[n] = 1;
		node start;
		start.x = n, start.step = 0;
		Q.push(start);
		bfs();
	}
	return 0;
}