Catch That Cow POJ

题目链接:点我

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    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.

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题意:

你每次可以达到当前点的左边或者右边或者当前点2倍坐标的位置,问是否能达到目标.

思路:

简单的bfs,同时标记你走过的位置就可以了.

代码:

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;struct ss{    int x,step;}q[100000+10];bool vis[100000+10];int n,k;void bfs(){    memset(vis,false,sizeof(vis));    int head=0;    int tail=0;    q[tail].x=n;    q[tail++].step=0;    vis[n]=true;    while(head<tail){        ss p=q[head++];        if(p.x==k){            cout<<p.step<<endl;            return ;        }        if(p.x-1>=0&&vis[p.x-1]==false){            q[tail].x=p.x-1;            q[tail++].step=p.step+1;            vis[p.x-1]=true;        } if(p.x+1<=100000&&vis[p.x+1]==false){            q[tail].x=p.x+1;            q[tail++].step=p.step+1;            vis[p.x+1]=true;        } if(p.x*2 <= 100000&&vis[p.x*2]==false){            q[tail].x=p.x*2;            q[tail++].step=p.step+1;            vis[p.x*2]=true;        }    }}int main(){    while(scanf("%d %d",&n,&k)!=EOF){        bfs();    }    return 0;}