Catch That Cow (BFS)_catch that cow bfs-优快云博客

本文介绍了一种在数轴上通过不同移动方式寻找特定目标的算法。利用BFS（宽度优先搜索）策略，探讨了如何从起点出发，在最短的时间内达到目标点，通过实例演示了算法的具体实现。

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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

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：

在数轴上从n移动到k，有3种移动的方式，设当前位置的坐标为x，则一步可以到达x-1，x+1，2x的位置，

问最少经过几步可以到达指定位置

代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int MAX = 100005*2;
struct node{
	int x;
	int step;
};
int vis[MAX];
int n,k;
void Bfs(){
	struct node tem,temp;
	queue<node>p;
	temp.x=n;
	temp.step=0;
	vis[n]=1;
	p.push(temp);
	while(!p.empty()){
		tem=p.front();
		p.pop();
		if(tem.x==k){
			printf("%d\n",tem.step);
			return;
		}
		int x=tem.x,step=tem.step;
		//枚举3种方式 
		if(x-1>=0&&!vis[x-1]){ //左移动1位 
			vis[x-1]=1;
			temp.x=x-1;
			temp.step=step+1;
			p.push(temp);
		}
		if(x<k&&!vis[x+1]){ //右移动一位 
			vis[x+1]=1;
			temp.x=x+1;
			temp.step=step+1;
			p.push(temp);
		}
		if(x<k&&!vis[x*2]){ //向右2倍的位置 
			vis[x*2]=1;
			temp.x=x*2;
			temp.step=step+1;
			p.push(temp);
		}
	}
}
int main(){
	scanf("%d%d",&n,&k);
	Bfs();
	return 0;
}