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?

###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:
农场主的牛不见了,主人和牛在一条直线上,且牛没有新的目标,它不会走动,主人的位置是你n,牛的位置是k,主人可以有三种走路的方法,右左(距离+-1),闪现(距离+x,x为当前位置),每走一步,一分钟,问几分钟主人能找到牛。bfs搜索方向即为三个“方向”。搜索所有走法;

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#include"iostream"
#include<queue>
#include"string.h"
using namespace std;

int n,k;
bool sign[200007];

struct node{
int x,step;
};

bool check(int a)
{
if(!sign[a]&&a>=0&&a<110000)
return true;
return false;
}

void bfs()
{
node u,v;
queue<node> q;
v.x=n;//初始化起点
v.step=0;
q.push(v);
sign[v.x]=true;
while(!q.empty()){
u=q.front();
q.pop();
if(u.x==k){
cout<<u.step<<endl;
return ;
}

//三种前进方向,左右和闪现
v=u;
v.x++;
v.step++;
if(check(v.x)){
sign[v.x]=true;
q.push(v);
}

v=u;
v.x--;
v.step++;
if(check(v.x)){
sign[v.x]=true;
q.push(v);
}

v=u;
v.x=2*v.x;
v.step++;
if(check(v.x)){
sign[v.x]=true;
q.push(v);
}
}
}

int main()
{
cin>>n>>k;
memset(sign,0,sizeof(sign));
bfs();
return 0;
}
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