Catch That Cow

Time Limit: 2000MSMemory Limit: 65536KB

SubmitStatistic

ProblemDescription

Farmer John has been informedof the location of a fugitive cow and wants to catch her immediately. He startsat 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 oftransportation: walking and teleporting. * Walking: FJ can move from any pointX to the points X - 1 or X + 1 in a single minute * Teleporting: FJ can movefrom any point X to the point 2 × X in a single minute. If the cow, unaware ofits pursuit, does not move at all, how long does it take for Farmer John toretrieve it?

Input

Line 1: Two space-separatedintegers: N and K

Output

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

ExampleInput

5 17

ExampleOutput

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 thefollowing path: 5-10-9-18-17, which takes 4 minutes.

Author

#include <iostream>
#include<bits/stdc++.h>

using namespace std;
struct node
{
int x,y,step;
}st;
int n,m,flag,book[200000];
int bfs()
{
    queue<node>g;
    memset(book,0,sizeof(book));
    book[st.x] = 1;
    g.push(st);
    while(!g.empty())
    {
        node now = g.front();
        g.pop();
        if(now.x==m)
        return now.step;
        for(int j=0;j<3;j++)
        {
            node next;
            if(j==0)
            {
            next.x = now.x+1;
            }
            else if(j==1)
            {
            next.x = now.x-1;
            }
            else if(j==2)
            {
            next.x = now.x*2;
            }
            next.step = now.step+1;
            if(next.x==m)
            {
                return next.step;
            }
            if(next.x>=0&&next.x<=200000&&!book[next.x])
            {
                book[next.x] = 1;
                g.push(next);
            }

        }

    }
    return 0 ;

}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        st.x = n;
        st.step = 0;
        printf("%d\n",bfs());
    }
    return 0;
}


/***************************************************
User name: jk160505徐红博
Result: Accepted
Take time: 4ms
Take Memory: 944KB
Submit time: 2017-02-15 21:00:28
****************************************************/