SDUT_1028_Catch That Cow(bfs)_01108845211776351727073110swh0700x2002-优快云博客

本文介绍了一个有趣的算法问题——如何在数轴上以最少步骤从一个起点到达另一个终点，可以通过步行或传送的方式移动。提供了详细的解决方案及代码实现。

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

题意：在一个数轴上，人在N位置，牛在K位置，人去找牛（人从 N到K），牛不动，人有两种走法，一种是+1或-1，另一种是在x位置，直接跳到2*x位置。（说明x 到x+1,x-1,2*x是连通的）
示例是： 5 17
可以从5跳到10，再从10到9，从9跳到18,从18到17，走4步；
也可以 5->4，4->8,8->16,16->17 都是4 步，用的步数最少；
分析：
当人的位置小于牛的位置，即n < k 时，可以进行+ 、-、 * ，因为通过这些都可以增大，而 n > k时只能进行减，这样才能找到牛。
当找到牛的时候，输出所用的步数。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int book[500005];

struct node
{
    int x;
    int s;
}que[500005];
int main()
{
    int n,k;
    int head = 1;
    int tail = 1;
    scanf("%d%d",&n,&k);

    que[head].x = que[tail].x = n;
    que[tail++].s = 0;
    book[n] = 1;
    while(head < tail)
    {
        int m = que[head].x;
        //找到牛
        if(m == k)
        {
            printf("%d\n",que[head].s);
            break;
        }
        //人的位置小于牛的
        if(m <= k&&book[m-1]==0)
        {
            book[m-1]=1;
            que[tail].x = m-1;
            que[tail++].s = que[head].s+1;
        }
        if(m <= k && book[m+1]==0)
        {
            book[m+1]=1;
            que[tail].x = m+1;
            que[tail++].s = que[head].s+1;
        }
        if(m <= k&&book[m*2]==0)
        {
            book[m*2] = 1;
            que[tail].x = m*2;
            que[tail++].s = que[head].s+1;
        }
        //人的位置大于牛
        if(m > k && book[m-1]==0)
        {
            book[m-1]=1;
            que[tail].x = m-1;
            que[tail++].s = que[head].s+1;
        }
        head ++ ;
    }
    return 0;
}