HDU 2727 Catch That Cow

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转载最新推荐文章于 2025-09-07 20:32:33 发布 · 72 阅读

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原文链接：http://www.cnblogs.com/GODLIKEING/p/3283600.html

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#数据结构与算法 #java

本文介绍了一道名为CatchThatCow的问题，该问题要求使用BFS算法找到农夫约翰抓住逃跑奶牛所需的最短时间。文章给出了具体实现代码，并通过剪枝优化算法效率。

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5457 Accepted Submission(s): 1740

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

    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

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teddy

思路：BFS + 剪枝，我是一坑货，剪枝只会皮毛

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
using namespace std ;
int n ,k ;
int map [ 100010 ];
struct Node
{
    int x ;
    int step ;
    bool operator < ( const Node &t ) const
     {
           return t .step < step ;
     }
}info ;
int BFS ()
{
    priority_queue < Node > q ;
    info .x = n ;
    info .step = 0 ;
    q .push (info );
    map [info .x ] = 1 ;
    while (!q .empty ())
    {
        Node boss ;
        boss = q .top ();
        q .pop ();
        if (boss .x == k )
            return boss .step ;
        int a = boss .x ;
        int step = boss .step ;
        //printf("%d              %d\n",a,step);
        if (a * 2 <= 100000 && a < k && map [a * 2 ] == 0 )
        {
                info .x = a * 2 ;
                info .step = step + 1 ;
                q .push (info );
                map [info .x ] = 1 ;
                //printf("%d   %d\n",info.x,info.step);
                if (info .x == k )
                    return info .step ;
        }
        if (a - 1 >= 0 && map [a - 1 ] == 0 )
        {
                info .x = a - 1 ;
                info .step = step + 1 ;
                q .push (info );
                map [info .x ] = 1 ;
                //printf("%d   %d\n",info.x,info.step);
                if (info .x == k )
                    return info .step ;
        }
        if (a + 1 <= 100000 && map [a + 1 ] == 0 )
        {
                info .x = a + 1 ;
                info .step = step + 1 ;
                q .push (info );
                map [info .x ] = 1 ;
                //printf("%d   %d\n",info.x,info.step);
                if (info .x == k )
                    return info .step ;
        }
    }
}
int main ()
{
        while (~scanf ( "%d%d" ,&n ,&k ))
        {
              memset (map , 0 , sizeof (map ));
              printf ( "%d\n" ,BFS ());
        }
}

转载于:https://www.cnblogs.com/GODLIKEING/p/3283600.html