POJ - 3278 Catch That Cow（BFS）

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?
InputLine 1: Two space-separated integers: N and KOutputLine 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.

思路：两种情况：1.人在牛右边，这种情况很明显，只能一次一次-1
                            2.人在牛左边，这边用bfs来做，只有三种情况（-1，+1，*2）。然后就可以了。

代码如下：#include<cstdio>
#include<queue>
#include<string.h>
using namespace std;

int f[100005];
struct node{
int x;
int step;
};

int bfs(int l,int r){
node st;
node next;
st.x=l;
st.step=0;
f[st.x]=1;
queue<node> q;
q.push(st);
while(!q.empty()){
st=q.front();
q.pop();
for(int i=0;i<3;i++){
if(i==0)
next.x=st.x+1;
else if(i==1)
next.x=st.x-1;
else if(i==2)
next.x=st.x*2;
//st.x+=next;
if(next.x<0||next.x>=100010)
continue;
if(!f[next.x])
{
f[next.x]=1;
next.step=st.step+1;
q.push(next);
if(next.x==r)
return next.step;
}

}
}

}

int main(){
int n,k;
while(~scanf("%d%d",&n,&k)){
memset(f,0,sizeof(f));
if(n>=k)
printf("%d\n",n-k);
else
printf("%d\n",bfs(n,k));
}
return 0;
}