hdoj-2717-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.

一个人在追一个宠物，有两种方式，步行：每秒前进或后退1步,借助传输工具：每秒前进2*x步；
直接bfs就好了

#include<cstdio>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

int vis[100005];
int x,ex,n,k;
struct go
{
    int x,step;
};

int bfs()
{
    queue<go>q;
    while(!q.empty()) q.pop();
    go next,now;
    now.step=0;
    now.x=n;
    vis[n]=1;
    q.push(now);
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        for(int i=1;i<=3;i++)
        {
            if(i==1) next.x=now.x-1;
            else if(i==2) next.x=now.x+1;
            else if(i==3) next.x=now.x*2;
            next.step=now.step+1;
            if(next.x==k)
              return next.step;
              if(next.x>=0&&next.x<=100000&&!vis[next.x])
              {
                  q.push(next);
                  vis[next.x]=1;
              }
        }
    }
}

int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        x=n;
        ex=k;
        if(x>=ex)
        {
            printf("%d\n",x-ex);
            continue;
        }
        int ans=bfs();
        printf("%d\n",ans);
    }
    return 0;
}