Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21112 Accepted Submission(s): 6127
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
题目大意：一条数轴上，给出一个初始位置和要到达的位置，每次可以+1，-1，*2操作，问最少几次操作能从起始位置到达最终位置。
具体代码如下：
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int N=100000+5;
bool map[N];
struct node{
int x,time;
};
int k,n;
int check(int x)
{
if(x>=N||map[x]==true||x<0)
return 0;
return 1;
}
int BFS()
{
queue<node> q;
node now,next;
now.x=n;
now.time=0;
q.push(now);
while(!q.empty())
{
now=q.front();
q.pop();
if(now.x==k)
return now.time;
//1
next.x=now.x+1;
next.time=now.time+1;
if(check(next.x))
{
q.push(next);
map[next.x]=1;
}
//2
next.x=now.x-1;
next.time=now.time+1;
if(check(next.x))
{
q.push(next);
map[next.x]=1;
}
//3
next.x=now.x*2;
next.time = now.time + 1;
if(check(next.x))
{
q.push(next);
map[next.x]=1;
}
}
return -1;
}
int main()
{
int ans;
while(cin>>n>>k)
{
memset(map,0,sizeof(map));
ans=BFS();
cout<<ans<<endl;
}
return 0;
}