poj3278 BFS入门

M - 搜索

Crawling in process... Crawling failed Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3278

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.

题目大意：农夫去追他跑丢的牛，题目给出了他和牛的位置，用数字N和K表示，假定牛不动，问农夫移动到牛的位置的最小步数，农夫每次的移动有三种选择：位置加1，位置减1，位置乘2.

思路分析：求最小步数，用BFS即可水过，不过由于本弱BFS刚刚入门，在做题的时候还是出现了很多问题，BFS一般要采用队列来进行实现，刚开始使用的是queue<int>队列，但是在对步数的保存上出现了问题，这时候我选择了queue<pair <int,int> >来记录位置和步数，每次将n+1,n-1,n*2都压入到队列当中去，直到n==m，结束。程序可以运行，样例的输出答案也是正确的，但是提交的时候MLE了，这让我明白了剪枝的重要性，最后程序在两个方面进行了剪枝，一个是建立一个标记数组，对搜索到的每一点都进行标记，避免重复搜索，第二就是对于n>m的情况,毫无疑问应该采取n-1。应用了这两点剪枝策略以后，成功A掉。

代码：#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;
const int maxn=4e5+5;
int n,m;
queue<pair<int,int> >q;
bool hash[maxn];
int main()
{
    int m,n;
  while(cin>>n>>m)
  {
    memset(hash,false,sizeof(hash));
    pair<int,int> p;
    p.first=n;
    p.second=0;
    hash[n]=true;
    q.push(p);
    while(!q.empty())
    {
      pair<int,int> a,b;
      a=q.front();
      if(a.first==m)
      {
          cout<<a.second<<endl;
          break;
      }
     a.second++;
     b=a;
     if(b.first>m)
     {
         b.first=a.first-1;
         if(hash[b.first]==false)
         {
           hash[b.first]=true;//标记该点，表示已经走过
           q.push(b);
         }
     }
     if(b.first<m)
     {
         b.first=a.first+1;
         if(hash[b.first]==false)
         {
           hash[b.first]=true;//标记该点，表示已经走过
           q.push(b);
         }
         b.first=a.first*2;
        if(hash[b.first]==false)
         {
           hash[b.first]=true;
           q.push(b);
         }
         b.first=a.first-1;
         if(hash[b.first]==false)
         {
           hash[b.first]=true;
           q.push(b);
         }
     }
     q.pop();//弹出队列第一个元素
    }
    while(!q.empty())
        q.pop();//注意清空队列
  }
    return 0;
}

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