poj 3278 Catch That Cow bfs

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 70246 Accepted: 22095

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK

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

把每个状态搜一遍

记得特判

ACcode：

#include <map>#include <queue>#include <cmath>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define maxn 100100using namespace std;int n,k;struct Node{    int x;    int t;};bool vis[maxn];Node my,now,s1,s2,s3;bool judge(Node x){    if(x.x<0||x.x>100000||vis[x.x])return false;    return true;}void bfs(){    if(n>=k){        printf("%d\n",n-k);        return;    }    queue<Node>q;    memset(vis,0,sizeof(vis));    my.x=n;    my.t=0;    q.push(my);    vis[n]=1;    while(!q.empty()){        now=q.front();q.pop();        s1=now;s1.t++;s1.x+=1;        if(judge(s1)){            if(s1.x==k){                printf("%d\n",s1.t);                return;            }else {                vis[s1.x]=1;                q.push(s1);            }        }        s2=now;s2.t++;s2.x-=1;        if(judge(s2)){            if(s2.x==k){                printf("%d\n",s2.t);                return;            }else {                vis[s2.x]=1;                q.push(s2);            }        }        s3=now;s3.t++;s3.x*=2;        if(judge(s3)){            if(s3.x==k){                printf("%d\n",s3.t);                return;            }else {                vis[s3.x]=1;                q.push(s3);            }        }    }}int main(){    while(~scanf("%d%d",&n,&k))bfs();    return 0;}