POJ 3278：Catch That Cow-优快云博客

本文链接：https://blog.youkuaiyun.com/u010885899/article/details/47037603

本文探讨了使用广度优先搜索算法解决寻找逃逸奶牛问题的过程，通过分析奶牛和农夫的位置关系，利用农夫的行走和瞬移两种方式，最小化追捕时间。实例演示了广度优先搜索在路径寻找问题中的有效应用。

Catch That Cow

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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.

发现广度优先搜索适合什么题呢？就是那种在每个点都给你几个选择，然后问你最短路径的问题，对，就是这样，这样的题目最适合广度优先搜索。

这次的这个就是，每次Farmer有三个选择，然后求最短到达目的地的行程。简直是广搜的模板题。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <queue>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int color[1000005];
int dis[1000005];

queue<int> q;

int main()
{
	//freopen("i.txt","r",stdin);
	//freopen("o.txt","w",stdout);

	int N,K;
	cin>>N>>K;
	
	if(N==K)
	{
		cout<<0<<endl;
	}
	else
	{
		memset(color,0,sizeof(color));
		memset(dis,0,sizeof(dis));

		q.push(N);
		while(!q.empty())
		{
			N=q.front();
			q.pop();
			if(N-1==K || N+1==K || 2*N==K)
			{
				cout<<dis[N]+1<<endl;
				break;
			}

			if(color[N-1]==0 && N-1>=0)
			{
				color[N-1]=1;
				dis[N-1]=dis[N]+1;
				q.push(N-1);
			}
			if(color[N+1]==0)
			{
				color[N+1]=1;
				dis[N+1]=dis[N]+1;
				q.push(N+1);
			}
			if(color[2*N]==0 && 2*N<=100000)
			{
				color[2*N]=1;
				dis[2*N]=dis[N]+1;
				q.push(2*N);
			}
		}
	}
	return 0;
}