E - Catch That Cow
POJ - 3278
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.
题意:JOHN要抓住他的牛,有两种移动方式,坐标加一或减一,坐标翻倍,牛不会动,求最少移动几次。
解法:宽搜,递归,开一个和范围一样大的数组来记录坐标是否到达过,从起始点进行三种移动,将移动后坐标标记并加到队列中,如果移动后坐标已经被标记了则返回,这说明一定可以用更少的移动到达这一点。
代码:
#include
#include
using namespace std;
int ans=0;
int a[100100]={0,};
int n,k;
struct point{
int x,step;
};
queue q;
point start;
int BFS()
{
while(!q.empty()){
point now;
now=q.front();
q.pop();
if(now.x==k){
n=k;
ans=now.step;
return ans;
}
for(int i=0;i<3;i++){
point next;
if(i==0)
next.x=now.x+1;
else if(i==1)
next.x=now.x-1;
else
next.x=now.x*2;
next.step=now.step+1;
if(next.x>=0&&next.x<100100&&a[next.x]==0){
a[next.x]=1;
q.push(next);
}
}
}
return ans;
}
int main()
{
cin>>n>>k;
start.x=n;
start.step=0;
q.push(start);
a[n]=1;
cout<