文章标题 POJ 3278 ： Catch That Cow（BFS）

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.
题意：John想尽快找到逃跑的牛，已知牛的位置k一直没有改变，而John每次可以往三个位置移动，当前位置为x时，下一次可以到x+1，x-1，x*2，问最少John最少需要多少时间到牛的位置。
分析：BFS，从x开始，往三个方向移动，如果是k就输出到达此点的时间，此时到达k时就是最短的时间，如果不是k就将此数放进队列。
代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<vector>#include<math.h>#include<map>#include<queue> #include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;int n,k; struct node {//结构体，位置和到达该位置的最短步数    int x,step;};node a[100005];int flag[100005];int bfs(int n){    for (int i=0;i<=100000;i++){        a[i].x=i;        a[i].step=inf;    }    memset(flag,0,sizeof (flag));    flag[n]=1;    a[n].x=n;    a[n].step=0;    queue <node> q;    q.push(a[n]);    int ans=0;    while (!q.empty()){        node tmp=q.front();        q.pop();        if (tmp.x==k){//判断是不是x是返回答案            ans=tmp.step;            return ans;        }        if (tmp.x-1>=0&&tmp.x-1<=100000&&!flag[tmp.x-1]){//判断是否在范围内，是的话就放进队列            node t; t.x=tmp.x-1; t.step=tmp.step+1; flag[tmp.x-1]=1;            q.push(t);        }        if (tmp.x+1>=0&&tmp.x+1<=100000&&!flag[tmp.x+1]){            node t; t.x=tmp.x+1; t.step=tmp.step+1; flag[tmp.x+1]=1;            q.push(t);        }        if (tmp.x*2>=0&&tmp.x*2<=100000&&!flag[tmp.x*2]){            node t; t.x=tmp.x*2; t.step=tmp.step+1; flag[tmp.x*2]=1;            q.push(t);        }    }    return -1;  }int main (){    while (scanf ("%d%d",&n,&k)!=EOF){        int ans=bfs(n);//得到答案        printf ("%d\n",ans);    }    return 0;}