poj3278 Catch That Cow

最新推荐文章于 2024-09-04 11:33:10 发布

原创最新推荐文章于 2024-09-04 11:33:10 发布 · 187 阅读

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本文介绍了一个简单的抓牛游戏算法，游戏中农夫通过行走和瞬间移动的方式，在最短的时间内捕捉到静止不动的牛。使用广度优先搜索(BFS)算法实现，详细展示了代码结构与实现过程。

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 94444		Accepted: 29624

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.

题目大意：告诉农夫和牛的初始位置，农夫要抓住牛，每次只能向左或者向右或者向乘以2的地方走，牛是不动的，问最少多少步能抓住

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
#define N 100000
int vis[N+10];//标记 
struct Step{ 
	int x;//存位置 
	int steps;//存步数 
	Step(int xx,int ss):x(xx),steps(ss){
	}
};

queue<Step> q;

int main(){
	int n,m;
	cin>>n>>m;
	memset(vis,0,sizeof(vis));
	q.push(Step(n,0));
	vis[n]=1;
	while(!q.empty()){//BFS 
		Step s = q.front();
		q.pop();
		if(s.x==m){
			cout<<s.steps<<endl;
			return 0;
		}
		else{
			if(s.x-1>=0&&!vis[s.x-1]){//向左 
				q.push(Step(s.x-1,s.steps+1));
				vis[s.x-1]=1;
			}
			if(s.x+1<=N&&!vis[s.x+1]){//向右 
				q.push(Step(s.x+1,s.steps+1));
				vis[s.x+1]=1;
			}
			if(s.x*2<=N&&!vis[s.x*2]){//*2 
				q.push(Step(s.x*2,s.steps+1));
				vis[s.x*2]=1;
			}
		}
	}
	return 0;
}