bfs入门练习Catch That Cow

对于bfs宽度搜索基础练习都是大同小异，也就是说，基本上都需要一个标记数组，加上一个队列，其本质就是不停的入队，出队，直到找到结果为止，BFS对于求最短路径等问题，具有很好地优势，但是同时要注意剪枝，也就是用标记数组跳过已经检索的部分。

题目：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 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.

题目大意为，在一条直线上给你两个数字，有3种移动数字的方式吗，一种是坐标+1，一种是-1，一种是*2，最后要求你输出，从第一个坐标到第二个坐标所花的最短时间，每次移动算一分钟。

分析：

这显然是一个广度搜索的问题，对于每一种情况所能到达的位置入队，然后不断更新，直到找到为止。

AC代码

#include<cstdio>#include<queue>#include<cstring>using namespace std;int book[1000000];int n,k;struct bu{int x;int s;};int check(int x){if(x<0||x>1000000||book[x]==1){return 0;}else return 1;}int bfs(int x){queue<bu> pi;bu p,q,xa;q.x=x;q.s=0;book[x]=1;pi.push(q);while(!pi.empty()){p=pi.front();pi.pop();if(p.x==k){return p.s;}xa.x=p.x+1;if(check(xa.x)){book[xa.x]=1;xa.s=p.s+1;pi.push(xa);}xa.x=p.x-1;if(check(xa.x)){book[xa.x]=1;xa.s=p.s+1;pi.push(xa);}xa.x=2*p.x;if(check(xa.x)){book[xa.x]=1;xa.s=p.s+1;pi.push(xa);}}return -1;}int main(){int sum;scanf("%d%d",&n,&k);sum=bfs(n);printf("%d",sum);}