SDUT-Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KB

Submit Statistic Discuss

Problem 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.

Example Input

5 17

Example Output

4

Hint

poj3278 有链接提示的题目请先去链接处提交程序，AC后提交到SDUTOJ中，以便查询存档。 
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.

题意：一条线上，农夫位于位置N，牛位于位置K，农夫有三种前进方式：1.N-1 2.N+1 3.N*2. 求出农夫抓到牛所用的最短时间

#include<bits/stdc++.h>using namespace std;const int ma=1000000;int n,k,Map[ma+10];///需要设立一个记录步数的变量struct node{    int step;    int data;} s1,s2;///此题不要用多个结构体变量解答,会超内存,用q2对三种情况分别判断queue<node>q;int check(int x){    if(x<0||x>=ma||Map[x]==1)    {        return 0;    }    else        return 1;}int BFS(int x){    s1.data=x;    s1.step=0;    Map[x]=1;//已走过该点    q.push(s1);    while(!q.empty())    {        s1=q.front();        q.pop();        if(s1.data==k)            return s1.step;        s2.data=s1.data+1;        if(check(s2.data))        {            s2.step=s1.step+1;            Map[s2.data]=1;            q.push(s2);        }        s2.data=s1.data-1;        if(check(s2.data))        {            s2.step=s1.step+1;            Map[s2.data]=1;            q.push(s2);        }        s2.data=s1.data*2;        if(check(s2.data))        {            s2.step=s1.step+1;            Map[s2.data]=1;            q.push(s2);        }    }    return -1;}int main(){    int ans;    while(scanf("%d%d",&n,&k)!=EOF)    {        memset(Map,0,sizeof(Map));        while(!q.empty()) q.pop();//注意调用前先清空，再就是不存在q.clear()        ans=BFS(n);        printf("%d\n",ans);    }    return 0;}