HDU2717 Catch That Cow BFS

最新推荐文章于 2021-02-28 08:15:18 发布

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本文介绍了一个基于搜索算法的问题——如何让农夫尽快抓到不动的逃逸奶牛。通过使用广度优先搜索策略，结合行走与瞬移两种移动方式，实现了最短时间内的捕获。代码示例详细展示了算法实现过程。

水题

当n<k 时三种操作 +1 -1 *2

当k>n时一种操作 -1 （减少无效状态）

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4481 Accepted Submission(s): 1426

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

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.

Source

USACO 2007 Open Silver

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>

using namespace std;

int n,k;
bool vis[100010];
struct node
{
    int x,cnt;
};


bool jud(int x)
{
    if( x>100005 || x<0 || vis[x])
        return false;
    return true;
}

int bfs(int s,int e)
{
    queue<node> q;
    node a,k;
    a.x=s,a.cnt=0;
    q.push(a);vis[s]=1;
    while(!q.empty())
    {
        a=q.front();
        q.pop();
        k.cnt=a.cnt+1;
        k.x=a.x-1;
        if(jud(k.x))
        {
            if(k.x==e) return k.cnt;
            q.push(k);
            vis[k.x]=1;
        }
        if(a.x<e)
        {
            k.x=a.x+1;
            if(jud(k.x))
            {
                if(k.x==e) return k.cnt;
                q.push(k);
                vis[k.x]=1;
            }
            k.x=a.x*2;
            if(jud(k.x))
            {
                if(k.x==e) return k.cnt;
                q.push(k);
                vis[k.x]=1;
            }
        }
    }
}

int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        memset(vis,0,sizeof(vis));
        if(k==n) printf("0\n");
        else printf("%d\n",bfs(n,k));
    }
    return 0;
}