POJ--3278 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

题意：额，大概意思就是农场主要去找他自己的牛，农场主在N行，牛在K行，农场主一次可以走+1，-1和2*N这三种行进路线，每一种都会花费1分钟，那么问题来了，农场主最快要几分钟能够捉到自己的牛。

注意：要注意特殊数据的考虑比如0-100000或者100000-0一道比较正常的广搜。

ac代码：

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
#define INF 0xfffffff
int vis[100010];
int k,n,cn[2]={1,-1},ans;
struct node{int n,step;
}a,temp;
int jud(int x){if(x>=0&&x<=100000&&vis[x]==0)return 1;return 0;
}
void bfs(int x){queue<node>q;a.n=x;a.step=0;q.push(a);vis[a.n]=1;int i;while(!q.empty()){a=q.front();q.pop();for(i=0;i<3;i++){if(i==2)temp.n=a.n*2;elsetemp.n=a.n+cn[i];temp.step=a.step+1;if(jud(temp.n)){vis[temp.n]=1;if(temp.n==k){if(ans>temp.step)ans=temp.step;    continue ;} q.push(temp);}}}}int main(){while(scanf("%d%d",&n,&k)!=EOF){memset(vis,0,sizeof(vis));ans=INF;if(n>=k)printf("%d\n",n-k);else{bfs(n);printf("%d\n",ans);    }}return 0;
}

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