kuangbin专题一 简单搜索 C - Catch That Cow

C - 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.

水题。。。。数组开大点不然会Runtime Error。。。。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int t[1000005],N,K;
		int sum;
void bfs()
{
		queue <int> x;
		x.push(N);
		t[N]=1;
		while(!x.empty()){
			int temp=x.front();

			x.pop();
			if(temp==K){
				sum=t[temp]-1;
				cout<<sum<<endl;
				return ;
			}
			if(!t[temp-1]&&(temp-1)>0){
				t[temp-1]=t[temp]+1;
				x.push(temp-1);
			}
			if(!t[temp+1]&&(temp+1)<=100000){
				t[temp+1]=t[temp]+1;
				x.push(temp+1);
			}
			
			if(!t[temp*2]&&(temp*2)<=100000){
				t[temp*2]=t[temp]+1;
				x.push(2*temp);
			}
		}
		return;
}
int main() 
{
	while(cin>>N>>K){
		if(N>K||N==K){
			cout<<N-K<<endl;
			continue;
		} 
		memset(t,0,sizeof(t));
		bfs();
	}
  return 0;
}