hdu 2717 (一维广搜）

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14027    Accepted Submission(s): 4293

Problem 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<iostream>#include<queue>#include<memory.h>using namespace std;int n,m;int vis[100005]={0};int dir[2]={-1,1};struct node{int x;int step;}s,e;void bfs(){queue<node>q;node x1,t;s.step=0;vis[s.x]=1;q.push(s);while(!q.empty()){t=q.front();//cout<<t.x<<endl;q.pop();if(t.x==e.x){cout<<t.step<<endl;return;}else{for(int i=0;i<2;i++){x1.x=t.x+dir[i];if(x1.x>=0&&x1.x<=100000&&vis[x1.x]==0){x1.step=t.step+1;vis[x1.x]=1;q.push(x1);}}x1.x=t.x*2;if(x1.x>=0&&x1.x<=100000){x1.step=t.step+1;vis[x1.x]=1;q.push(x1);}}}cout<<-1<<endl;}int main(){    while(cin>>n>>m)    {     memset(vis,0,sizeof(vis));    s.x=n;    e.x=m;    bfs();}}