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.
题意：有一个农夫在N位置，他得牛在K位置。他每一步可以向左走一步，向右走一步，走到当前位置的2倍处。问农夫碰到牛所需的最短步骤。
思路：BFS三个状态，初始为N的位置。边界判断！！！可行判断！！！
代码如下：
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
struct Node{
int id;
int cont;
};
int n,k;
int visit[200005]; //虽然没看到题意说每个位置只能走一次，但是不标记的话会内存超限
void bfs(){
queue <Node> Q;
while(!Q.empty())
Q.pop();
Node h;
h.id = n;
h.cont = 0;
visit[n] = 1;
Q.push(h);
while(!Q.empty()){
Node res = Q.front();
Q.pop();
if(res.id == k){
cout<<res.cont<<endl;
break;
}
int ID = res.id;
int Cont = res.cont;
if(ID <= k && !visit[ID+1]){ //可行判断
visit[ID+1] = 1;
Node temp;
temp.id = ID+1;
temp.cont = Cont+1;
Q.push(temp);
}
if(ID >= 1 && !visit[ID-1]){ //可行判断
visit[ID-1] = 1;
Node temp;
temp.id = ID-1;
temp.cont = Cont+1;
Q.push(temp);
}
if(ID <= k && !visit[ID*2]){ //可行判断
visit[ID*2] = 1;
Node temp;
temp.id = ID*2;
temp.cont = Cont+1;
Q.push(temp);
}
}
}
int main()
{
while(cin>>n>>k){
memset(visit,0,sizeof(visit));
bfs();
}
return 0;
}