POJ 3278 Catch That Cow/信息奥赛一本通 1253：抓住那头牛

Catch That Cow

Description

Farmer John has been informed of thelocation of a fugitive cow and wants to catch her immediately. He starts at apoint N (0 ≤ N ≤ 100,000) on a number lineand the cow is at a point K (0 ≤ K ≤ 100,000)on the same number line. Farmer John has two modes of transportation: walkingand teleporting.

* Walking: FJ can move from any point X tothe points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X ina single minute.

If the cow, unaware of its pursuit, doesnot move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Twospace-separated integers: N and K

Output

Line 1: The leastamount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

Hint

The fastest wayfor 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​​

算法分析：

题意:

农夫知道一头牛的位置，想要抓住它。农夫和牛都位于数轴上，农夫起始位于点N(0<=N<=100000)，牛位于点K(0<=K<=100000)。农夫有两种移动方式：

1、从X移动到X-1或X+1，每次移动花费一分钟

2、从X移动到2*X，每次移动花费一分钟

分析：bfs思想。

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>  
#include<list>  
using namespace std;  
const double eps = 1e-8;  
typedef long long LL;  
typedef unsigned long long ULL;  
const int INT_INF = 0x3f3f3f3f;  
const int INT_M_INF = 0x7f7f7f7f;  
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;  
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;  
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};  
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};  
const int MOD = 1e9 + 7;  
const double pi = acos(-1.0);  
const int MAXN = 1e5 + 10;  
const int MAXT = 10000 + 10;  
#define N 1000000
int n,m,vis[N];   
int bfs()  
{  
     memset(vis,0,sizeof(vis));
     queue<int>q;
     
 
     q.push(n);
     int t=0;
     while(!q.empty())
     {
        int t=q.front();
        q.pop();
        if(t==m)
        {
            printf("%d",vis[t]);
            break;
        }
        //三种路径
        if(t>0&&!vis[t-1])     
        {
            q.push(t-1);
            vis[t-1]=vis[t]+1;
        }
        if(t<m&&!vis[t+1])
        {
            q.push(t+1);
            vis[t+1]=vis[t]+1;
        }
        if(2*t<N&&!vis[2*t])//注意此时的t*2的范围
        {
            q.push(2*t);
            vis[2*t]=vis[t]+1;
        }
     }
      return 0;
}  
int main()  
{  
      
    while(scanf("%d%d",&n,&m)!=EOF)  
    {  
        if(n>m)  //巧妙剪枝，在农夫前面直接往前找最快
        {
            printf("%d\n",n-m);
            continue;
        }
        bfs();
        
    }  
    return 0;  
      
}