Catch That Cow

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

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.

思路：

bfs将情况都遍历出来，找到最先的就行了。（多组数据不然会wa）

#include <iostream>
#include <queue>
using namespace std;
const int maxn = 100010;
struct NODE {
	int x;
	int time;
};
int bfs(int n, int m) {
	bool book[maxn] = {false};
	queue<NODE> q;
	NODE top;
	top.time = 0;
	top.x = n;
	q.push(top);
	while (!q.empty()) {
		NODE now = q.front();  
		q.pop();
		if (now.x == m) return now.time;
		for (int i = 0; i < 3; i++) {
			int next = 0; 
			if (i == 0) next = now.x + 1;
			else if (i == 1) next = now.x - 1;
			else next = now.x * 2;
			if (next < 0 || next > maxn || book[next]) continue;
			book[next] = true;
			top.time = now.time + 1;
			top.x = next;
			q.push(top);
		}
	}
	return -1;
}
int main() {
	int n, m;
	while (scanf("%d %d", &n, &m) != EOF) {
		if (n >= m) printf("%d\n", n - m);
		else printf("%d\n", bfs(n, m));
	}
	return 0;
}