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Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK

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.

代码：

#include <cstdio>
#include <queue>

using namespace std;

const int MAXN = 100001;

int N, K;

int vis[MAXN];
int ret[MAXN];

queue<int> q;

int BFS(int s, int d)
{
 if (s == d) return 0;
 q.push(s);

 int cur;

 while (!q.empty())
 {
  cur = q.front();
  q.pop();
  if (cur + 1 < MAXN && !vis[cur + 1])
  {
   q.push(cur + 1);
   ret[cur + 1] = ret[cur] + 1;
   vis[cur + 1] = 1;
  }

  if (cur + 1 == d) break;

  if (cur - 1 >= 0 && !vis[cur - 1])
  {
   q.push(cur - 1);
   ret[cur - 1] = ret[cur] + 1;
   vis[cur - 1] = 1;
  }

  if (cur - 1 == d) break;

  if (cur << 1 < MAXN && !vis[cur << 1])
  {
   q.push(cur << 1);
   ret[cur << 1] = ret[cur] + 1;
   vis[cur << 1] = 1;
  }

  if (cur << 1 == d) break;
 };

 return ret[d];
}

int main()
{
 scanf("%d %d", &N, &K);
 printf("%d\n", BFS(N, K));
 return 0;
}

分析：

在判断是否越界的时候，不能像题目中所说，当牧场主所在的位置大于10W的时候，就认为他越界。   因为他有可能先去到

100010的时候 ，在回来。所以再判断是时候，越界的最大值最好为20W。这样就不会出错了。