Catch That Cow (BFS广搜)
问题描述:
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
Output
Sample Input
5 17
Sample Output
4
Hint
解题思路:
代码:
#include<cstdio>
#include<queue>
#include<cstring>
#include<iostream>
using namespace std;
int n,k;
int a[]; struct node
{
int x;
int step;
}now,net; int bfs(int x)
{
queue<node> q;
now.x=x;
now.step=;
q.push(now);
while(q.size())
{
now=q.front();
q.pop();
//三种情况
if(now.x==k)return now.step;
net.x=now.x+;
if(net.x>=&&net.x<=&&a[net.x]==)
{
a[net.x]=;
net.step=now.step+;
q.push(net);
}
net.x=now.x-;
if(net.x>=&&net.x<=&&a[net.x]==)
{
a[net.x]=;
net.step=now.step+;
q.push(net);
}
net.x=now.x*;
if(net.x>=&&net.x<=&&a[net.x]==)
{
a[net.x]=;
net.step=now.step+;
q.push(net);
}
}
return -;
} int main()
{
while(~scanf("%d%d",&n,&k))
{
memset(a,,sizeof(a));
a[n]=;
int ans=bfs(n);
printf("%d\n",ans);
}
return ;
}
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