POJ3278 Catch That Cow —— BFS

题目链接：http://poj.org/problem?id=3278

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 97563		Accepted: 30638

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

题解：

一开始以为是数学找规律，但是看到数据范围很小，n<=1e5，可以用数组存；而且题目要求的是“最少步数”，而“最少步数”经常都是用BFS求的。再将BFS的思想带入看看，发现可以解决问题。

在第一次提交时，RUNTIME ERROR了。但是再看看代码，数组没有开小，不是数组的问题。后来发现再判断某个位置是否vis时，先判断了是否vis，然后再判断这个位置是否合法，这样会导致数组溢出，所以问题就出现在这里了，需谨慎！！

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e5+;

 int vis[MAXN];

 struct node
 {
     int val, step;
 };

 queue<node>que;
 int bfs(int n, int k)
 {
     ms(vis,);
     while(!que.empty()) que.pop();

     node now, tmp;
     now.val = n;
     now.step = ;
     vis[n] = ;
     que.push(now);

     while(!que.empty())
     {
         now = que.front();
         que.pop();

         if(now.val==k)
             return now.step;

         tmp.step = now.step+;
         if(now.val+>= && now.val+<=1e5 && !vis[now.val+] )  //先判断范围再判断vis ！！！
             vis[now.val+] = , tmp.val = now.val+, que.push(tmp);
         if(now.val->= && now.val-<=1e5 && !vis[now.val-] )
             vis[now.val-] = , tmp.val = now.val-, que.push(tmp);
         if(now.val*>= && now.val*<=1e5 && !vis[now.val*] )
             vis[now.val*] = , tmp.val = now.val*, que.push(tmp);
     }
 }

 int main()
 {
     int n, k;
     scanf("%d%d",&n, &k);
     cout<< bfs(n,k) <<endl;
 }