HDU2717 Catch That Cow （BFS）

Catch That Cow

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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.        

#include <cstdio>#include <cstring>#include <queue>#define N 100010using namespace std;int map[N];struct node{int x;int step;};int Judge(int x){if(x < 0 || x > N || map[x])return 0;return 1;}int bfs(int n, int k){queue <node> Q;node temp,a;a.x = n;a.step = 0;map[n] = 1;Q.push(a);while(!Q.empty()){a = Q.front();Q.pop();if(a.x == k)return a.step;temp = a;temp.x = a.x + 1;if(Judge(temp.x)){temp.step = a.step + 1;map[temp.x] = 1;Q.push(temp);}temp.x = a.x - 1;if(Judge(temp.x)){temp.step = a.step + 1;map[temp.x] = 1;Q.push(temp);}temp.x = 2 * a.x;if(Judge(temp.x)){temp.step = a.step + 1;map[temp.x] = 1;Q.push(temp);}}}int main(){int n,k;while(~scanf("%d%d",&n,&k)){memset(map,0,sizeof(map));int ans = bfs(n,k);printf("%d\n",ans);}return 0;}