HDU2717 Catch That Cow (BFS)

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

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.

经典BFS。对所有情况进行一次搜索，最先找到的就是解。

代码如下：

#include <iostream>#include <queue>#include <cstring>using namespace std;const int N=1000000;int map[N+10],n,k;struct point{int x,step;} ; int check(int x){ if(x<0||x>N||map[x]) { return 0; }return 1;} int bfs(int x){queue<point> q;  //队列 point now,next;now.x=x;now.step=0; map[x]=1;  //标记 q.push(now); //将now插入队列while(!q.empty()){now=q.front(); //将先进入队列的元素取出存入nowq.pop(); //取出后删除if(now.x==k) return now.step;next=now;next.x=now.x+1;//三种情况 if(check(next.x)){ next.step=now.step+1; //步数加一； map[next.x]=1;//标记 q.push(next);}next.x=now.x-1;if(check(next.x)){next.step=now.step+1;map[next.x]=1;q.push(next);}next.x=now.x*2;if(check(next.x)){next.step=now.step+1;map[next.x]=1;q.push(next);} } return -1; } int main(){scanf("%d%d",&n,&k);memset(map,0,sizeof(map));//标记为0（表示未搜索过）cout<<bfs(n)<<endl;return 0; }

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