POJ-3278-Catch That Cow

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.

这是一道简单的BFS题，只是将以前的4个方向变成两个方向就行。同时这里需要注意开vis数组的时候一定要开大些，不然会一直出现RE

代码如下：

#include <cstdio>#include <queue>#include <algorithm>#include <cstring>typedef long long ll;using namespace std;ll maxn = 200005;ll n,k;ll vis[200005];struct node{    ll x,time;};bool check(ll x){    if(vis[x])        return true;    else if(x<0)        return true;    return false;}ll bfs(){    node s,next;    s.x = n;    s.time = 0;    vis[s.x]=1;    queue<node>q;    q.push(s);    while(!q.empty())    {        s = q.front();        q.pop();        if(s.x==k)            return s.time;        for(int a = 0; a < 3; a ++)        {            next = s;            if(a==0)                next.x = s.x+1;            else if(a==1)                next.x = s.x-1;            else if(a==2&&s.x<k)                next.x = s.x*2;            if(check(next.x))                continue;            next.time = s.time+1;            vis[next.x]=1;            q.push(next);        }    }    return 0;}int main(){    scanf("%lld%lld",&n,&k);        memset(vis,0,sizeof(vis));        ll ans = bfs();        printf("%lld",ans);    return 0;}