Catch That Cow广搜典型例题_(1)catch that cow 题目描述:farmer john has been inform-优快云博客

本文介绍了一个关于利用广度优先搜索(BFS)算法解决寻找最小时间以从起点到达目标点的问题。具体场景设定为农夫约翰通过步行或瞬移的方式在数轴上追赶一只静止不动的奶牛。

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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

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.

/*
	对于本题，一眼看出用广搜。
		因为是线性的（没有太多元素），所以不必使用结构体 
		首先将标志数组初始化为0； 
		接下来便是构造队列，构造队列的作用是：使多个线索同时前进，符合广搜的要求，
		也就是说，广搜最核心的东西便是队列的循环；
		在循环中 ，务必使每一次前进方式都要涉及到，这样才能广搜到最佳路径： 
		在大部分广搜题中 ，都设置一个结构体，以及结构体类型的队列，在结构体成员中设置要求的结果变量 	
*/ 
#include<iostream>
#include<algorithm>
#include<memory.h>
#include<cstdio> 
#include<queue>
using namespace std;
const int MAXN=100001;
const int M=0; 
int n,k,vis[MAXN];
void bfs(){
	queue<int> op; 
	op.push(n);
	while(!op.empty()) {
		int fir=op.front();
		op.pop(); 
		if(fir==k) return ; 
		if(fir+1<MAXN&&!vis[fir+1]) {
			vis[fir+1]=vis[fir]+1; 
			op.push(fir+1); 
		} 
		if(fir-1>=0&&!vis[fir-1]) {
			vis[fir-1]=vis[fir]+1; 
			op.push(fir-1); 
		} 
		if(fir*2<MAXN&&!vis[fir*2]) {
			vis[fir*2]=vis[fir]+1; 
			op.push(fir*2); 
		}	
	}  
} 

int main()
{

	while(~scanf("%d%d",&n,&k))
	{
		memset(vis,0,sizeof(vis)); 
		bfs();
		cout<<vis[k]<<endl; 
	} 
	return 0; 
} 
//修改了6个小时才AC，水了一逼阿