POJ 3278 (第一道队列题,爽)

POJ 3278 (第一道队列题,爽)

学会了队列.这道题的主要思想:

5先入列   把5出列  5 可以变成 4 ,6 , 10, ,把得到的数入列,然后再出列分别处理..如果前面出现过就不入列..定义一个计数数组,然后number[y - 1] = number[y] + 1;  number[y + 1] = number[y] + 1; number[y * 2] = number[y] + 1;

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8341		Accepted: 2476

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 Code

Problem: 3278		User: luoguangyao
Memory: 1048K		Time: 110MS
Language: C++		Result: Accepted

• Source Code

  #include <iostream>
      #include <queue>
      using namespace::std;
      int number[100001] = {0};
      bool num[100001] = {0};
      int main()
      {
      queue<int> x;
      int a;
      int b;
      scanf("%d%d",&a,&b);
      int count = 0;
      number[a] = 0;
      x.push(a);
      while (x.size())
      {
      int y = x.front();
      x.pop();
      num[y] = 1;
      if (y == b)
      {
      break;
      }
      else
      {
      if (y - 1 >= 0)
      {
      if (!num[y - 1])
      {
      x.push(y - 1);
      number[y - 1] = number[y] + 1;
      num[y - 1] = 1;
      }
      }
      if (y + 1 <= 100000)
      {
      if (!num[y + 1])
      {
      x.push(y + 1);
      number[y + 1] = number[y] + 1;
      num[y + 1] = 1;
      }
      }
      if (y * 2 <= 100000)
      {
      if (!num[y * 2])
      {
      x.push(y * 2);
      number[y * 2] = number[y] + 1;
      num[y * 2] = 1;
      }
      }
      }
      }
      cout << number[b] << endl;
      return 0;
      }