[kuangbin]搜索 POJ 3278 Catch That Cow（bfs）

题目链接

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 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.

 

题目大意：给两个点，一个初始位置，一个要去的位置，有三种走的方法：+1，-1，*2，求最小步数

思路：广搜，以初始位置为起点就那么一搜的就行，，，，我的代码很好懂的

AC代码

#include 
#include 
#include 
using namespace std;
struct node{
	int x,step;
};
queueq;
int vis[100010];
void bfs(int n,int k){
	node now,nex;
	now.x=n;
	now.step=0;
	q.push(now);
	while(!q.empty()){
		now=q.front();
		q.pop();
		if(now.x==k){
			printf("%d\n",now.step);
			return;
		}
		for(int i=0;i<3;i++){
			switch(i){
				case 0:{
					nex.x=now.x+1;
					if(nex.x>100000) continue;
					break;
				}
				case 1:{
					nex.x=now.x-1;
					if(nex.x<0) continue;
					break;
				}
				case 2:{
					nex.x=now.x*2;
					if(nex.x>100000) continue;
					break;
				}	
			}
			if(vis[nex.x]==0){
				nex.step=now.step+1;
				q.push(nex);
				vis[nex.x]=1;
			}
		}
	}
}
int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	while(!q.empty()) q.pop();
	bfs(n,k);
	return 0;
}