HDU-2717 BFS

Problem 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
4
题意：
求X轴上n->k的最短步骤，（n->n-1,n->n+1,n->2*n）
解题思路：
BFS
代码：

#include<bits/stdc++.h>
using namespace std;
struct state{int x,step;
};
int vis[200001];
int bfs(int n,int k){memset(vis,0,sizeof(vis));state str,now,next;str.x=n,str.step=0;vis[n]=1;int tp;queue<state>q;q.push(str);while(!q.empty()){now=q.front(),q.pop();if(now.x==k)return now.step;for(int i=0;i<3;++i){if(i==0) tp=now.x-1;else if(i==1) tp=now.x+1;else tp=(now.x<<1);if(vis[tp]||tp<0||tp>100000) continue;vis[tp]=1;next.x=tp;next.step=now.step+1;q.push(next);}}return -1;
}
int main(){int n,k;while(~scanf("%d%d",&n,&k)){if(n>=k)printf("%d\n",n-k);else printf("%d\n",bfs(n,k));}return 0;
}

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