C++广度优先搜索算法之抓住那头牛(Catch that cow)

抓住那头牛:

农夫知道一头牛的位置，想要抓住它。农夫和牛都位于数轴上，农夫起始位于点N(0<=N<=100000)，牛位于点K(0<=K<=100000)。农夫有两种移动方式：

1、从X移动到X-1或X+1，每次移动花费一分钟

2、从X移动到2*X，每次移动花费一分钟

假设牛没有意识到农夫的行动，站在原地不动。农夫最少要花多少时间才能抓住牛？

Catch that cow：

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (0 ≤N ≤ 100,000) on a number line and the cow is at a pointK (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 pointX to the pointsX- 1 orX + 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?

以上是题目，具体输入输出如下：

输入

两个整数，N和K。

输出

一个整数，农夫抓到牛所要花费的最小分钟数。

样例输入 5 17 样例输出 4

代码如下:

#include
const int sb=1e6;
bool mark[sb+1];
int que[sb+1],pre[sb+1],n,k;
int next;
void print(int x)
{
	int num=0;
	while(pre[x])
	{
		num++;
		x=pre[x];
	}
	printf("%d",num);
}
void bfs()
{
	if(n==k){printf("0");return ;}
	int head=0,tail=1;
	que[1]=n;
	mark[n]=true;
	while(head!=tail)
	{
		head++;
		for(int i=0;i<3;i++)
		{
			switch(i)
			{
				case 0: next=que[head]+1;break;
				case 1: next=que[head]-1;break;
				case 2: next=que[head]*2;break;
			}
			if(next>=0&&next<=sb&&mark[next]!=true)
			{
				tail++;
				que[tail]=next;
				mark[next]=true;
				pre[tail]=head;
				if(next==k) 
				{
					print(tail);
					head=tail;
					break;
				}
			}
		}
	}
}
int main()
{
	scanf("%d%d",&n,&k);
	bfs();
}