hdu 2717 Catch That Cow bfs搜索 解题报告

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11107    Accepted Submission(s): 3447

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

 

分析：只能+1，-1，*2三种方向，跟hdu1548基本一样，直接拿了1548的代码改了几行。

#include <cstdio>
#include <iostream>
#include <queue>
#include <algorithm>
#include <cstring>
#include <stack>
#include <cmath>
using namespace std;

int n,a,b,vis[100010];
struct zc
{
    int now,step;
};

int check(zc a)
{
    if(a.now<0||a.now>100000)return 0;
    return 1;
}

int bfs()
{
    zc t,p;
    t.step=0;
    t.now=a;
    queue<zc> Q;
    Q.push(t);

    while(!Q.empty())
    {
        t=Q.front();
        Q.pop();
        if(t.now==b)return t.step;
        //+1
        p.now=t.now+1;
        p.step=t.step+1;
        if(check(p)&&!vis[p.now])
        {
            Q.push(p);
            vis[p.now]=1;
        }
        //-1
        p.now=t.now-1;
        p.step=t.step+1;
        if(check(p)&&!vis[p.now])
        {
            Q.push(p);
            vis[p.now]=1;
        }
        //*2
        p.now=t.now*2;
        p.step=t.step+1;
        if(check(p)&&!vis[p.now])
        {
            Q.push(p);
            vis[p.now]=1;
        }
    }
    return -1;
}

int main()
{
    while(scanf("%d %d",&a,&b)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        printf("%d\n",bfs());
    }
    return 0;
}