Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 43851 | Accepted: 13674 |
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.
Source
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
struct Node{
int x,t;
Node(int X,int T){
x=X; t=T;
}
};
int sign[200010]; //注意这里的标记数组要定大一些，定为100010会runtime error
int ans;
queue <Node> q;
int main(){
int n,k;
while(cin>>n>>k){
ans=9999999;
memset(sign,0,sizeof(sign));
while(!q.empty())
q.pop(); //清空队列
int x,t;
x=n; t=0;
q.push(Node(x,t)); //入队
sign[x]=1; //标记已访问
while(!q.empty()){
x=q.front().x; t=q.front().t;
q.pop();
if(x==k)
ans=min(ans,t);
else{
if(x<k){
if(sign[x+1]==0){
q.push(Node(x+1,t+1));
sign[x+1]=1;
}
if(sign[2*x]==0){
q.push(Node(2*x,t+1));
sign[2*x]=1;
}
if(x>0&&sign[x-1]==0){
q.push(Node(x-1,t+1));
sign[x-1]=1;
}
}
else if(x>k){
if(sign[x-1]==0){
q.push(Node(x-1,t+1));
sign[x-1]=1;
}
}
}
}
cout<<ans<<endl;
}
return 0;
}