Catch That Cow

题目

     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.



* 这道题是bfs的模板题 也就是求类似于最短路径的问题
* 如果我们用dfs去做的话就会TLE，因为我们只用求出最短的用时而不是把所有的到达方式都求出来
* 很坑的是这题是多输入的

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#include <iostream>
#include <queue>
#include <string.h>
#define MAX 100000
using namespace std;

queue<int> pos;
int t[MAX];
int vis[MAX];

int main() {
    int n,k,front;

    while(cin>>n>>k){
        memset(t,0,sizeof(t));
        memset(vis,0,sizeof(vis));
        pos = queue<int>();
        pos.push(n);
        vis[n] = 1;
        while(!pos.empty()){
           front = pos.front();
           if(front == k)
               break;
           if(front+1 <= k && !vis[front+1]){
               pos.push(front+1);
               t[front+1] = t[front]+1;
               vis[front+1] = 1;
           }
           if(front-1>=0 && !vis[front-1]){
               pos.push(front-1);
               t[front-1] = t[front]+1;
               vis[front-1] = 1;
           }
           if(front*2<=MAX && !vis[front*2]){
               pos.push(front*2);
               t[front*2] = t[front]+1;
               vis[front*2] = 1;
           }
           pos.pop();
        }
        cout<<t[k]<<endl;
    }
    return 0;
}