poj 3278 广搜

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 70333 Accepted: 22115

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：就是通过三种方式从一个点到另一个点

这里使用队列可以减少空间的使用，用数组模拟就不知道要多大的空间了

注意这里搜索 5 到时 17 是用了4 步，

但是17到 5  就会用 12 步，注意题目说的运动顺序

a[m-1]=a[m]+1;

上面这一步意思是：用了多少步走到点 m-1。其他同理可以得到

#include<stdio.h>#include<queue>#include<string.h>using namespace std;const int maxn=100001;int used[maxn];int a[maxn];int main(){    int m,n;    while(scanf("%d%d",&m,&n)!=EOF)    {        memset(used,0,sizeof(used));        memset(a,0,sizeof(a));        queue<int>q;        q.push(m);        used[m]=1;        while(q.size()!=0)        {            m=q.front();            q.pop();            if(m==n)                break;            if(m-1>=0&&used[m-1]==0){                q.push(m-1);                used[m-1]=1;                a[m-1]=a[m]+1;            }            if(m+1<=maxn&&used[m+1]==0){                q.push(m+1);                used[m+1]=1;                a[m+1]=a[m]+1;            }            if(m*2<=maxn&&used[m*2]==0){                q.push(m*2);                used[m*2]=1;                a[m*2]=a[m]+1;            }        }        printf("%d\n",a[n]);    }    return 0;}