Hrbust-1012-Catch That Cow（广搜BFS入门）

Catch That Cow
Time Limit: 2000 MS Memory Limit: 65536 K
Total Submit: 1993(563 users) Total Accepted: 819(513 users) Rating: Special Judge: No
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
For each test case :

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

Process to the end of file.

Output
For each test case :

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

Sample Input
5 17
100 100
Sample Output
4
0

题意：在一维数轴中，输入初始位置，和目的位置，通过 前进一步（坐标+1），后退一步（坐标-1），跳跃（坐标位置*2），三种方式到达目的位置，输出最小移动步骤

广度优先搜索入门

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
int main()
{
    int st,k;
    queue<int>cow;//建立队列
    int vis[100009];//声明标记数组
    while(scanf("%d%d",&st,&k)!=EOF)//输入初始位置和目标位置
    {
        memset(vis,0,sizeof(vis));//清空标记数组
        while(!cow.empty())//清空队列，若不为空，则弹出队列内元素
        {
                cow.pop();
        }
        cow.push(st);//将初始位置放入队列
        vis[st]=1;//标记初始位置为1
        while(!cow.empty())//开始循环走步骤，条件是队列不为空
        {
            int walk=cow.front();//每次走一步都要记录队首的坐标值，记录队首坐标值以确定你下一步的行走方式
            cow.pop();//出队列
            if(walk-1>=0&&!vis[walk-1])//后退一步
            {
                cow.push(walk-1);
                vis[walk-1]=vis[walk]+1;//这里vis不仅标记了每个坐标是否被走过，同时也记录了已经行走的步数
            }
            if(walk+1<=100000&&!vis[walk+1])//前进一步
            {
                cow.push(walk+1);
                vis[walk+1]=vis[walk]+1;
            }
            if(walk*2<=100000&&!vis[walk*2])//跳跃
            {
                cow.push(walk*2);
                vis[walk*2]=vis[walk]+1;
            }
            if(vis[k])//若目的位置被标记，则结束循环
            {
                break;
            }
        }
        printf("%d\n",vis[k]-1);//输出步数，并减掉初始位置标记的多余的一步
    }
    return 0;
}