Catch That Cow

Catch That Cow

Time Limit:2000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX+ 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 andK

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. 

题意：人在N的位置，牛在K的位置，人有三种前进方式，+1、-1、*2，求人找到牛的最少的步数

#include"queue"#include"cstdio"#include"cstring"#include"iostream"#include"algorithm"using namespace std;#define MAXN 100000struct node{    int loc,t;};int st,en;bool vis[MAXN + 5];queue < node > q;int bfs(int st){    while(!q.empty())    {        q.pop();    }    node a;    a.loc = st;    a.t = 0;    vis[a.loc] = true;    q.push(a);    while(!q.empty())    {        node b = q.front();        q.pop();        if(b.loc == en)        {            return b.t;        }        int l;        l = b.loc + 1; //情况1，前进一步        if(l > 0 && l < MAXN && !vis[l])        {            node c;            c.loc = l;            c.t = b.t + 1;            vis[l] = true;            q.push(c);        }        l = b.loc - 1; //情况2，后退一步        if(l >= 0 && l <= MAXN && !vis[l])        {//因为是后退，所以l可以等于0，之前写的>0，那么1,0这组样例就会出错            node c;            c.loc = l;            c.t = b.t + 1;            vis[l] = true;            q.push(c);        }        l = b.loc << 1;        if(l >= 0 && l <= MAXN && !vis[l]) //情况3，*2        {            node c;            c.loc = l;            c.t = b.t + 1;            vis[l] = true;            q.push(c);        }    }}int main(){    while(~scanf("%d%d",&st,&en))    {        memset(vis,false,sizeof(vis)); //BFS被访问过的就不会再次被访问，需要标记        printf("%d\n",bfs(st));    }    return 0;}