poj 3278:Catch That Cow（简单一维广搜）

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 45648		Accepted: 14310

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

---

 

　　简单一维广搜。

　　题意：

　　你在一个一维坐标的n位置，牛在k位置，你要从n到k，抓到那头牛。你可以有三种走法，n+1，n-1，或者n*2直接跳。求你抓到那头牛的最短步数。

　　思路：

　　简单广搜的思想。状态跳转的时候有三种跳转的方式，将新的状态放到队列中，再不断提取队列中最前面的状态，直到找到k位置。

　　代码：

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <queue>
 using namespace std;

 bool isw[];

 struct Node{
     int x;
     int s;
 };

 bool judge(int x)
 {
     if(x< || x>)
         return true;
     if(isw[x])
         return true;
     return false;
 }

 int bfs(int sta,int end)
 {
     queue <Node> q;
     Node cur,next;
     cur.x = sta;
     cur.s = ;
     isw[cur.x] = true;
     q.push(cur);
     while(!q.empty()){
         cur = q.front();
         q.pop();
         if(cur.x==end)
             return cur.s;
         //前后一个个走
         int nx;
         nx = cur.x+;
         if(!judge(nx)){
             next.x = nx;
             next.s = cur.s + ;
             isw[next.x] = true;
             q.push(next);
         }
         nx = cur.x-;
         if(!judge(nx)){
             next.x = nx;
             next.s = cur.s + ;
             isw[next.x] = true;
             q.push(next);
         }
         //向前跳
         nx = cur.x*;
         if(!judge(nx)){
             next.x = nx;
             next.s = cur.s + ;
             isw[next.x] = true;
             q.push(next);
         }
     }
     return ;
 }

 int main()
 {
     int n,k;
     while(scanf("%d%d",&n,&k)!=EOF){
         memset(isw,,sizeof(isw));
         int step = bfs(n,k);
         printf("%d\n",step);
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013