POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 98011 Accepted: 30779

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.

思路：构图然后广度优先搜索。

#include<cstdio>#include<cstring>using namespace std;#define MAX 100005int font[2*MAX];//存储所到位置的坐标int hash[2*MAX];//记录到下标坐标所走的步数（最大位置是100000*2）int n,k;int has(int i){    if(i>=2*MAX||i<0||hash[i])        return 0;//剪枝，对之前判断过的点使程序不再对其作处理    else return 1;}int BFS(int n,int k){    if(n==k)return 0;//牛和农夫处在同一坐标    int x,first=0,last=0;    font[first++]=n;    while(last<first)    {        //printf("first=%d last=%d\n",first,last);        x=font[last++];        //printf("x=%d hash[%d]=%d\n",x,x,hash[x]);        if(x-1==k||x+1==k||x*2==k)            return hash[x]+1;//特殊坐标，一步即可完成        if(has(x-1)==1)            hash[x-1]=hash[x]+1,font[first++]=x-1;        if(has(x+1)==1)            hash[x+1]=hash[x]+1,font[first++]=x+1;        if(has(x*2)==1)            hash[x*2]=hash[x]+1,font[first++]=x*2;    }//三个if把三种情况都走一遍并做记录。    return -1;}int main(){    while(scanf("%d %d",&n,&k)!=EOF)    {        memset(hash,0,sizeof(hash));        int a=BFS(n,k);        printf("%d\n",a);    }    return 0;}