POJ 3278 Catch That Cow (附有Ｒuntime Error和Wrong Answer的常见原因）

题目链接：http://poj.org/problem?id=3278

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 124528		Accepted: 38768

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：给你两个点 N，K，你有三种走法：X+1 ，X-1，Ｘ*2　，求从Ｎ到Ｋ走的最少步数。

代码附有Wrong Answer 和　Runtime Error的几个参考原因。

 #include<iostream>
 #include<queue>
 #include<cstring>
 #include<cstdio>
 using namespace std;

 /**
  * 造成Runtime Error的原因有两点：
  * 1.数组开小了，也就是下面的maxn可能只开到 1e5稍多一点，这是不对的，在搜索过程中有个 2*x ,这会导致数组大小应该大于1e5的两倍。
  * 2.在判断!mark[x-1]的时候，没有x>0这一个约束条件，如果x = 0 ，则 x-1会导致数组越界；
  * 3.在 x+1 和 2*x这两种情况下，应该限制 x<=K;
  */
  /**
   * 造成Wrong Answer 的原因之一：在判断 x-1 的情况的时候，不要有 x<=K
   */
 const int maxn = 2e5+;
 bool mark[maxn];
 int N,K;
 struct node
 {
     int x;
     int step;
 };

 //队列的写法;
 void bfs()
 {
     queue<node> q;
     struct node cu,ne;
     cu.x = N;
     cu.step = ;
     mark[N] = ;
     q.push(cu);
     while(!q.empty())
     {
         cu = q.front();
         q.pop();
         if(cu.x == K)
         {
             printf("%d\n",cu.step);
             return ;
         }
         //下面的三种情况应该是并列关系，不应该满足一种情况就不判断别情况了
         //我开始写成了if-else if-else if形式，想成不是并列的形式了；
         //如果实在找不到哪里错了，可以打印cu.x变量，根据变量的值找问题；
         if(cu.x>&&!mark[cu.x-])
         {
             ne.x = cu.x - ;
             ne.step = cu.step + ;
             mark[ne.x] = ;
             q.push(ne);
         }
         if(cu.x<=K&&!mark[cu.x+])
         {
             ne.x = cu.x + ;
             ne.step = cu.step + ;
             mark[ne.x] = ;
             q.push(ne);
         }
         if(cu.x<=K&&!mark[cu.x*])
         {
             ne.x = *cu.x;
             ne.step = cu.step + ;
             mark[ne.x] = ;
             q.push(ne);
         }
     }
 }
 int main()
 {
     while(scanf("%d%d",&N,&K)!=EOF)
     {
         memset(mark,,sizeof(mark));
         bfs();
     }
     return ;
 }

Ac代码