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最新推荐文章于 2024-11-01 00:16:00 发布

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农夫约翰得知一头逃犯奶牛的位置，他从点N出发，目标是点K，两者都在数轴上。约翰有两种移动方式：步行和瞬移。步行可以向左或向右移动1单位，瞬移可以将位置变为原来的2倍。如果奶牛不动，约翰如何最快抓住它？输入包含两个整数N和K，输出约翰抓到奶牛的最短时间。例如，当N=5, K=17时，最短路径5-10-9-18-17需要4分钟。" 132932429,20036702,SDIO总线在音视频设备中的应用解析,"['音视频开发', '接口标准', '数据传输', '嵌入式硬件']

深搜广搜的队列写法模板

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.

简历模版把坐标放入队列，三种情况依次进队列

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
int book[1000010];
int n,k;
struct node
{
    int x,step;
};
int cmp(int x)
{
    if(x<0 || x>=1000000 || book[x])
        return 0;
    return 1;
}
int bfs(int x)
{
    int i;
    queue<node> Q;
    node a,next;
    a.x = x;
    a.step = 0;
    book[x] = 1;
    Q.push(a);
    while(!Q.empty())
    {
        a = Q.front();
        Q.pop();
        if(a.x == k)
            return a.step;
        next = a;
        next.x = a.x+1;
        if(cmp(next.x))
        {
            next.step = a.step+1;
            book[next.x] = 1;
            Q.push(next);
        }
        next.x = a.x-1;
        if(cmp(next.x))
        {
            next.step = a.step+1;
            book[next.x] = 1;
            Q.push(next);
        }
        next.x = a.x*2;
        if(cmp(next.x))
        {
            next.step = a.step+1;
            book[next.x] = 1;
            Q.push(next);
        }
    }
    return -1;
}
 
int main()
{
    int ans;
    while(~scanf("%d%d",&n,&k))
    {
        memset(book,0,sizeof(book));
        ans = bfs(n);
        printf("%d\n",ans);
    }
    return 0;
}