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?
5 17
4
/*
广度优先搜索,重要的是剪枝和标记访问过的路径。
所谓的剪枝,也可以理解为生成每一个子节点的条件;所谓条件限制越严谨,剪枝效果越好。
*/
#include
#include
#include
using namespace std;
const int MAX=100020; // 这个MAX的设定并不太严谨
int visited[MAX]={0};
int main()
{
int n,k;
while (cin>>n>>k){
if( n > k ){
cout< q;
q.push(n);
int t=0;
while( !q.empty() ){
t = q.front();
q.pop();
if(t == k){
break;
}
if( t < k && !visited[t+1] ){ // 当t>k时,不执行+1 action
q.push(t+1);
visited[t+1] = visited[t] + 1;
}
if( t > 0 && !visited[t-1] ){ //
q.push(t-1);
visited[t-1] = visited[t] + 1;
}
if( 2*t < MAX && !visited[2*t] ){ //此处不太严谨,如上讨论
q.push(2*t);
visited[2*t] = visited[t] +1;
}
}
cout<