Catch That Cow 寻找那头牛

J - Catch That Cow
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status

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.

这道题是采用广度优先搜索和用队列来存放数据。

广度优先搜索是分层，起点是0 层，然后第一个分之是1 层一次往下寻找，

先把起点的位置和步数放在队列里，然后再访问第二层······················

//广度优先搜索，用队列来存放。

#include <stdio.h>

#include <string.h>

#include <algorithm>

#include <iostream>

#include <queue>

using namespace std;

const int MAXN =100000; // c++中定义一个常数，而在c中则是用#define MAXN 100000；

int visited[MAXN+10]; //设一个数组来记录节点是否被走过，（查重） 为真时表示走过这个点

struct step //建立一个step的一个结构体

{

int x; //农夫的位置

int steps; //所走的步数

step(int xx,int s):x(xx),steps(s) {} //是一个构造函数：是在声明变量时调用的 。。。

//:x(xx),steps(s)简单写法，将成员变量x初始化为参数xx，成员变量steps初始化为参数s；

//也可以写成：step(int xx=0,int s=0){this->x=xx;this->steps=s;}

};

queue<step> q; //建立一个step结构体形式的队列。

int main()

{

int n,k;

cin>>n>>k;

memset(visited,0,sizeof(visited)); //初始化为0，为1时则表示已走过。

q.push(step (n,0)); //将起始位置先放入队列里n,0代表steps所走的步数。

visited
=1; //将起始位置记录

while(!q.empty()) //

{

step s=q.front(); //将队列的头打出。放在close里

if(s.x==k) //如果遇到终节点则结束。

{

cout<<s.steps<<endl;

return 0;

}

else

{

if(s.x-1>=0&&!visited[s.x-1]) //向左走一步

{

q.push(step(s.x-1,s.steps+1));

visited[s.x-1]=1;

}

if(s.x+1<=MAXN&&!visited[s.x+1]) //向右走一步。

{

q.push(step(s.x+1,s.steps+1));

visited[s.x+1]=1;

}

if(s.x*2<=MAXN&&!visited[s.x*2]) //走的时原来的两倍位置。

{

q.push(step(s.x*2,s.steps+1));

visited[s.x*2]=1;

}

q.pop(); //弹出

}

}

return 0;

}