POJ3278BFS

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 94506 Accepted: 29650

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.

Source

USACO 2007 Open Silver

        题意：一维数轴上有两个点，要求求从一个点到另一个点的最小步数，在X可以走到X+1或X-1或2*X。

        解题思路：这个题是个很水的BFS题，基本套模板。以下是我的AC代码

#include <iostream>#include <queue>#include <cstdio>#include <memory.h>using namespace std;#define maxn 100005bool flag[maxn];struct trip{    int x,step;};int n,k;int check(int x){    if(x<0||x>=maxn) return 1;    return flag[x];}void bfs(){    queue<trip> q;    trip now,next;    now.x=n;    now.step=0;    flag[now.x]=true;    q.push(now);    while(!q.empty())    {        now=q.front();        q.pop();        if(now.x==k)        {            printf("%d\n",now.step);            break;        }        for(int i=0;i<3;i++)        {            if(i==0) next.x=now.x+1;            else if(i==1) next.x=now.x-1;            else if(i==2) next.x=now.x*2;            if(check(next.x)) continue;            next.step=now.step+1;            flag[next.x]=true;            q.push(next);        }    }}int main(){    //freopen("in.txt","r",stdin);    scanf("%d%d",&n,&k);    memset(flag,false,sizeof(flag));    bfs();    return 0;}