POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 44070 Accepted: 13764

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

简单的BFS应用，POJ刷题正式开始，poj计划的开始就意味着，我以后的生活就没以前这么轻松了。

送自己一句话吧：想玩好ACM，8个字：“题海战术！题海战术！”

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>#include <math.h>const int N =  10000010;using namespace std;int n,k;struct node{    int wz;    int ans;}q[N];int vis[N];int jz[] = {1,-1};void BFS(){    int s = 0,e = 0;    node f,t;    f.ans = 0;    f.wz = n;    q[e++] = f;    vis[n] = 1;    while(s<e)    {        t = q[s++];        if(t.wz==k)        {            printf("%d\n",t.ans);            return ;        }        for(int i = 0;i<3;i++)        {            if(i!=2)            {                f.wz = t.wz + jz[i];            }            else            {                f.wz = t.wz * 2;            }            if(vis[f.wz]==0 && f.wz>=0 && f.wz <=100000)            {                f.ans = t.ans + 1;                vis[f.wz] = 1;                q[e++] = f;            }        }    }}int main(){    while(~scanf("%d%d",&n,&k))    {        memset(vis,0,sizeof(vis));       BFS();    }    return 0;}