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

HintThe 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;
}