杭电 HDU 2717 Catch That Cow

http://acm.hdu.edu.cn/showproblem.php?pid=2717

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4938    Accepted Submission(s): 1566

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.
 

 

用的广搜：

AC代码：

#include<iostream>#include<cstdio>#include<queue>#include<cstring>using namespace std;int number[100010];struct Node{    int n;   //记录当前位置    int i;   //记录当前第几步};int Bfs(Node n, int k){    queue<Node> q;    Node now,next;    q.push(n);    while(!q.empty())    {        now = q.front();        q.pop();        number[now.n] = 2;        if(now.n != k)        {            next.i = now.i + 1;   //走的步数加一            next.n = now.n + 1;   //向右走一步            if(next.n >= 0 && next.n <= 100000 && number[next.n] == 0)            {                number[next.n] = 1;                q.push(next);            }            next.n = now.n - 1;   //向左走一步            if(next.n >= 0 && next.n <= 100000 && number[next.n] == 0)            {                number[next.n] = 1;                q.push(next);            }            next.n = now.n*2;     //跳到当前数字的两倍位置            if(next.n >= 0 && next.n <= 100000 && number[next.n] == 0)            {                number[next.n] = 1;                q.push(next);            }        }        else        {            return now.i;        }    }}int main(){    int n,k,num;    Node p;    while(scanf("%d%d",&n,&k)!=EOF)    {        memset(number,0,sizeof(number));        p.n = n;        p.i = 0;        num = Bfs(p,k);        printf("%d\n",num);    }    return 0;}