[HDOJ2717]Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8616    Accepted Submission(s): 2714

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.

 

经典BFS，常规思路。

注意：

题目所给5 17和17 5答案是不一样的。我曾天真地认为需要进行一步swap，实际上是不需要的。

遍历到某点应判断是否越界，否则也会ACCESS_VIOLATION。

数组也要开大一点，否则也会ACCESS_VIOLATION。

（PS：我忘记修改判断是否越界时界限的大小。）

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <queue>
 using namespace std;
 int vis[];
 int stp[];
 int n, k;

 void swap(int& a, int& b)
 {
     a = a ^ b;
     b = a ^ b;
     a = a ^ b;
 }

 /*
 int move(int sgn, int cur)
 {
     if(sgn == 1)
     {
         return cur + 1;
     }
     if(sgn == -1)
     {
         return cur - 1;
     }
     else
     {
         return cur * 2;
     }
 }
 */

 void BFS(int ini)
 {
     int cur, now = ;    //init
     queue<int> q;
     vis[ini] = ;
     stp[ini] = ;
     q.push(ini);
     while(!q.empty())
     {
         cur = q.front();
         q.pop();
         for(int i = ; i < ; i++)
         {
             if(i == )
             {
                 now = cur + ;
             }
             else if(i == )
             {
                 now = cur - ;
             }
             else
             {
                 now = cur * ;
             }

             if(!vis[now] && now >=  && now <= )
             {
                 vis[now] = ;
                 q.push(now);
                 stp[now] = stp[cur] + ;
             }
             if(now == k)
             {
                 printf("%d\n", stp[now]);
                 return ;
             }
         }
     }
 }
 int main()
 {
     while(scanf("%d %d", &n, &k) != EOF && n+k)
     {
         memset(vis, , sizeof(vis));
         memset(stp, , sizeof(stp));
         if (n == k)
         {
             printf("0\n");
         }
         else
         {
             BFS(n);
         }
     }
     return ;
 }