【HDU 2717 Catch That Cow】+ 队列_.12857 .cow-优快云博客

本文链接：https://blog.youkuaiyun.com/WYK1823376647/article/details/53177020

本文介绍了一道名为CatchThatCow的算法题目，该题描述了农夫约翰如何通过步行和传送两种方式，在最短时间内捕捉到静止不动的逃逸奶牛。文章提供了使用广度优先搜索(BFS)算法的C++实现代码。

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Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12857 Accepted Submission(s): 3976

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.

Sample Input

5 17

Sample Output

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.

DSF？队列似乎也可把～标记每个点～先进队列的肯定就是最优解吧～

AC代码：

#include<cstdio>
#include<queue>
using namespace std;
struct node{
   int x,y;
}v;
int a,b;
int DFS(){
    int vis[100010] = {0};
    queue <node> q;
    v.x = a,v.y = 0;
    q.push(v);
    while(!q.empty()){
        v = q.front();
        q.pop();
        if(v.x == b) return v.y;
        node w;w.x = v.x + 1; w.y = v.y + 1;
        if(w.x >= 0 && w.x <= 100000 && !vis[w.x])
            vis[w.x] = 1,q.push(w);
        w.x = v.x - 1 ; w.y = v.y + 1;
        if(w.x >= 0 && w.x <= 100000 && !vis[w.x])
            vis[w.x] = 1,q.push(w);
        w.x = v.x * 2 ; w.y = v.y + 1;
        if(w.x >= 0 && w.x <= 100000 && !vis[w.x])
            vis[w.x] = 1,q.push(w);
    }
}
int main()
{
    while(scanf("%d %d",&a,&b) != EOF){
        printf("%d\n",DFS());
    }
    return 0;
}