hdu 2717 Catch That Cow

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Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16120    Accepted Submission(s): 4823

Problem 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.

这是第一个bfs有思路的题，也是简单题，思路也没问题，但是一直是Runtime Erro,真搞不懂，后来看了看别人的，还是细节。细节，细节。

#include<cstdio>#include<queue>#include<cstring>using namespace std;const int N = 1000000;int vis[N+10];int n,k;struct node{    int x,step;};int check(int x){    if(x<0||x>=N||vis[x])        return 0;    return 1;}int bfs(int x){    queue<node>Q;    node a,next;    a.x=x;    a.step=0;    vis[x]=1;    Q.push(a);    while(!Q.empty())    {        a=Q.front();        Q.pop();        if(a.x==k)            return a.step;        next.x=a.x+1;        if(check(next.x))        {            next.step=a.step+1;            vis[next.x]=1;            Q.push(next);        }        next.x=a.x-1;        if(check(next.x))        {            next.step=a.step+1;            vis[next.x]=1;            Q.push(next);        }        next.x=a.x*2;        if(check(next.x))        {            next.step=a.step+1;            vis[next.x]=1;            Q.push(next);        }    }    return -1;}int main(){    while(~scanf("%d%d",&n,&k))    {        memset(vis,0,sizeof(vis));        int ans=bfs(n);        printf("%d\n",ans);    }    return 0;}