http://poj.org/problem?id=3278(bfs)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 76935		Accepted: 24323

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.

 

题意：给出两个数star，end，给出x-1，x+1，x*2四种运算，使star变成end;

 

思路：直接bfs就好了；

 

代码：

 #include <iostream>
 #include <string.h>
 #include <queue>
 #define MAXN 200000+10
 using namespace std;

 struct Node       //***用结构体表示节点，x表示当前值，step记录当前步数
 {
     int x, step;
 };

 int star, end, vis[MAXN];

 int bfs(int x)
 {
     Node a, b, c;
     a.x = x, a.step = ;      //***赋初值
     queue<Node>q;            //***用队列实现
     q.push(a);                //***头节点入队
     while(!q.empty())         //***如果队列不为空，继续循环
     {
         b = q.front();        //***取当前队列的头节点做父节点并将其从队列中删除
         q.pop();
         vis[b.x] = ;       //***标记
         if(b.x == end)       //***如果达到目标，返回步数
         {
             return b.step;
         }
         c = b;                  //***符合条件的儿子入队
         if(c.x >=  && !vis[c.x-])
         {
             c.x-=;
             c.step++;
             vis[c.x]=;
             q.push(c);
         }
         c = b;
         if(c.x < end && !vis[c.x+])
         {
             c.x+=;
             c.step++;
             vis[c.x]=;
             q.push(c);
         }
         c = b;
         if(c.x < end && !vis[*c.x])
         {
             c.x*=;
             c.step++;
             vis[c.x]=;
             q.push(c);
         }
     }
     return -;
 }

 int main(void)
 {
     while(cin >> star >> end)
     {
         if(star >= end)             //***如果star>=end，则每次减1，最少需要star-end步
         {
             cout << star - end << endl;
             continue;
         }
         memset(vis, , sizeof(vis));  //***标记数组清0，不然会对后面的测试产生影响
         int ans=bfs(star);            //***深搜
         cout << ans << endl;
     }
     return ;
 }