[POJ3278] Catch That Cow

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.

Source

USACO 2007 Open Silver

题意

在同一数轴上,有一头奶牛坐标为k,农夫的坐标为n,要从n运动到k。有三种方法：

1. 从当前位置x运动到x+1
2. 从当前位置x运动到x-1
3. 当前位置x运动到2*x

题解

裸裸的(BFS)。

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;

const int N=100001;
bool vis[N];
int n,k;
typedef pair<int,int> pr;

void Bfs()
{
	queue<pr> Q;
	Q.push(make_pair(n,0));
	vis[n]=1;
	while(!Q.empty())
	{
		if(Q.front().first==k)
		{
			printf("%d
",Q.front().second);
			return;
		}
		if((Q.front().first+1<N)&&(!vis[Q.front().first+1]))
		{
			vis[Q.front().first+1]=1;
			Q.push(make_pair(Q.front().first+1,Q.front().second+1));
		}
		if((Q.front().first-1>=0)&&(!vis[Q.front().first-1]))
		{
			vis[Q.front().first-1]=1;
			Q.push(make_pair(Q.front().first-1,Q.front().second+1));
		}
		if(((Q.front().first<<1)<N)&&(!vis[Q.front().first<<1]))
		{
			vis[Q.front().first<<1]=1;
			Q.push(make_pair(Q.front().first<<1,Q.front().second+1));
		}
		Q.pop();
	}
}

int main()
{
	scanf("%d%d",&n,&k),Bfs();
	return 0;
}