超超超简单的bfs——POJ-3278

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 89836		Accepted: 28175

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.

 

 

起点在n，终点是k，每一步可以是+1、-1或*2，最少走几步？

 

 #include<stdio.h>
 #include<queue>
 #include<string.h>
 using namespace std;
 int a[];//一个2倍大小的数组代表可以走到的位置，因为有乘二所以要开二倍以防RE，a[i]=x  -->  走到坐标i需要x步
 int main()
 {
     int n, k, t;
     queue<int>q;
     scanf("%d%d", &n, &k);
     memset(a, -, sizeof(a));//所有坐标初始化为-1
     a[n] = ;                //n走到n当然是需要步
     q.push(n);                //当前位点入队
     while (!q.empty())
     {
         t = q.front();        //读取队首
         q.pop();            //删除队首
         if (t == k)            //若到达终点则直接输出并结束
         {
             printf("%d\n", a[k]);
             return ;
         }
         if (t -  >=  && a[t - ] == -)//可能到达的结点入队，要判断是否越界，走过的不再走
         {
             a[t - ] = a[t] + ; q.push(t - );//下一步走到的位点所需步数是当前位点的步数+1
         }
         if (t +  <  && a[t + ] == -)
         {
             q.push(t + ); a[t + ] = a[t] + ;
         }
         if (t *  <  && a[t * ] == -)
         {
             q.push(t * ); a[t * ] = a[t] + ;
         }
     }
 }

简单的bfs