牛客 NC25080 Catch That Cow

题目描述

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?

输入描述:

Line 1: Two space-separated integers: N and K

输出描述:

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

示例1

输入

复制

5 17

输出

复制

4

说明

Farmer John starts at point 5 and the fugitive cow is at point 17.
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.

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.

思路：每次农夫可以向前一步、向后一步以及传送至i*2个位置，因为每次一步只会有三种情况，因此可以通过宽搜的方式每次第一次遍历到的点必定是最短路径；

AC代码：

#include
#include
#include
using namespace std;
 
const int N=100010,INF=100000;
 
int dist[N],q[N];
int n,k;
 
int bfs()
{
	int hh=0,tt=0;
	q[tt]=n;
	while(hh<=tt)
	{
		int t=q[hh++];
		if(t==k)return dist[k];
		if(t-1>=1&&!dist[t-1])
		{
			dist[t-1]=dist[t]+1;
			q[++tt]=t-1;
		}	
		if(t+1<=INF&&!dist[t+1])
		{
			dist[t+1]=dist[t]+1;
			q[++tt]=t+1;
		}
		if(t*2<=INF&&!dist[2*t])
		{
			dist[2*t]=dist[t]+1;
			q[++tt]=t*2;
		}
	}
	return -1;
}
 
int main()
{
	scanf("%d%d",&n,&k);
	cout<<bfs();
}