HDOJ 2717 Catch That Cow(BFS)

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8896    Accepted Submission(s): 2806

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.
 

 

ac代码：

#include<stdio.h>#include<iostream>#include<queue>#include<algorithm>#include<string.h>using namespace std;int n,m;int dir[2]={1,-1};int v[1000100];//一定要定义足够大struct s{int x;int step;}a,b;int check(int xx){if(xx<0||xx>=1000100||v[xx])//一定要把范围定义足够大，在两个范围问题上wrong了无数return 0;return 1;}int bfs(int w){int i;a.x=w;a.step=0;v[a.x]=1;queue<s>q;q.push(a);while(!q.empty()){a=q.front();q.pop();if(a.x==m){return a.step;}for(i=0;i<2;i++)//检查 +1，-1两种情况 {   b=a;   b.x=a.x+dir[i];   if(check(b.x))   {   b.step=a.step+1;   v[b.x]=1;//标记    q.push(b);   }    }b.x=a.x*2;//检查x2的情况 if(check(b.x)){b.step=a.step+1;v[b.x]=1;q.push(b);}}return 0;}int main(){while(scanf("%d%d",&n,&m)!=EOF){memset(v,0,sizeof(v));int num;num=bfs(n);printf("%d\n",num);}return 0; }