SDUT1028Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory limit: 65536K

题目描述

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?

输入

Line 1: Two space-separated integers: N and K

输出

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

示例输入

5 17

示例输出

4

提示

过程上图

示例程序

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <queue>#define Inf 100000using namespace std;struct node{    int data;    int step;};int vis[100000];int DFS(int n,int k){    queue<node>p;    node now,next;    now.data=n;    now.step=0;    vis[now.data]=1;    p.push(now);    while(!p.empty())    {        now=p.front();        p.pop();        next=now;        for(int i=0; i<3; i++)        {            if(i==0)                next.data=now.data-1;            else if(i==2)                next.data=now.data+1;            else                next.data=now.data*2;            next.step=now.step+1;            if(next.data==k)                return next.step;            if(next.data>=0&&next.data<=Inf&&!vis[next.data])            {                vis[next.data]=1;                p.push(next);            }        }    }    return 0;}int main(){    int n,k;    while(~scanf("%d%d",&n,&k))    {        memset(vis,0,sizeof(vis));        if(n<k)            printf("%d\n",DFS(n,k));        else if(n==k)            printf("0\n");        else            printf("%d\n",n-k);    }    return 0;}