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最新推荐文章于 2025-03-30 16:45:00 发布

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最新推荐文章于 2025-03-30 16:45:00 发布

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本文介绍了一道经典的算法题目——《CatchThatCow》，通过一维BFS算法解决农夫追赶静止不动的牛的问题。农夫有两种移动方式：步行和瞬间传送，最终目的是以最短时间抓住牛。

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Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 105873		Accepted: 33091

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

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：农夫在数轴上追牛，一次操作可以往前一步，也可以往后一步，也可以坐标乘2；

题解：一维BFS。

#include <iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
int n,k;
int vis[100005],step[100005];
int check(int n){
	if(vis[n])return 1;
	if(n>100000||n<0)return 1;
	return 0;
}
int bfs(){
	int t,next,i;
	queue<int>que;
	que.push(n);
	while(!que.empty()){
		t=que.front();
		que.pop();
		if(t==k)return step[t];
		for(i=1;i<=3;i++){
			if(i==1)next=t-1;
			if(i==2)next=t+1;
			if(i==3)next=2*t;
			if(check(next))continue;
			vis[next]=1;
			step[next]=step[t]+1;
			que.push(next);
		}
	}
}
int main()
{
	while(~scanf("%d%d",&n,&k)){
		memset(vis,0,sizeof(vis));
		memset(step,0,sizeof(step));
		printf("%d\n",bfs());

	}
    return 0;
}