原来这个叫做广搜啊。。。_广搜能不用队列吗-优快云博客

本文链接：https://blog.youkuaiyun.com/forpro_yang/article/details/6629833

本文解析了POJ3278题目——Catch That Cow的算法实现过程，详细介绍了如何利用广度优先搜索策略，在限定时间内找到最优路径以抓住逃逸的奶牛。文中还分享了使用C++实现的代码示例，包括如何运用STL queue库进行操作。

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题目来源 poj3278 ；

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536K

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 fastes 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.

这个题目的思想还是挺简单，也不要求剪枝优化什么的，所以根据题意直接写，另外，通过这个题目学了一些stl <queue>库函数的一些用法。。

代码：

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

int main()
{
	int line[100001]={0};
	bool sign[100001]={0};
	queue<int>x;        // 创建一个int型的队列。。
	int n,k;
	cin>>n>>k;
    	x.push(n);       // 将n压入队列x队首。
    	line[n]=0;       
	    while(!x.empty())
		{
		  int y=x.front()  ;     //将x的队头赋值给y。;
		  x.pop();              //上一步没有将队头取出，这一步取出队头。
		  sign[n]=true;         //初始位置设为走过。
		  if(y==k) break;        
		  else 
		  {
			if(y>=1&&!sign[y-1])
			 {
				x.push(y-1);             //将y-1压入x队首。
				line[y-1]=line[y]+1;
				sign[y-1]=true;
			 }
			if(y+1<=100000&&!sign[y+1])
			  {
				x.push(y+1);
				line[y+1]=line[y]+1;
				sign[y+1]=true;
			  }
			if(y<=50000&&!sign[y*2])
			  {
				x.push(y*2);
				line[y*2]=line[y]+1;
				sign[y*2]=true;
			  }
		  }
		}
		printf("%d\n",line[k]);

	
	return 0;
}