暑期第一弹<搜索> C - Catch That Cow（BFS）

C - Catch That Cow

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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.

题意：有一个农夫在N位置，他得牛在K位置。他每一步可以向左走一步，向右走一步，走到当前位置的2倍处。问农夫碰到牛所需的最短步骤。

思路：BFS三个状态，初始为N的位置。边界判断！！！可行判断！！！

代码如下：

#include <iostream>#include <queue>#include <cstring>using namespace std;struct Node{    int id;    int cont;};int n,k;int visit[200005];      //虽然没看到题意说每个位置只能走一次，但是不标记的话会内存超限void bfs(){    queue <Node> Q;    while(!Q.empty())        Q.pop();    Node h;    h.id = n;    h.cont = 0;    visit[n] = 1;    Q.push(h);    while(!Q.empty()){        Node res = Q.front();        Q.pop();        if(res.id == k){            cout<<res.cont<<endl;            break;        }        int ID = res.id;        int Cont = res.cont;        if(ID <= k && !visit[ID+1]){        //可行判断            visit[ID+1] = 1;            Node temp;            temp.id = ID+1;            temp.cont = Cont+1;            Q.push(temp);        }        if(ID >= 1 && !visit[ID-1]){        //可行判断            visit[ID-1] = 1;            Node temp;            temp.id = ID-1;            temp.cont = Cont+1;            Q.push(temp);        }        if(ID <= k && !visit[ID*2]){        //可行判断            visit[ID*2] = 1;            Node temp;            temp.id = ID*2;            temp.cont = Cont+1;            Q.push(temp);        }    }}int main(){    while(cin>>n>>k){        memset(visit,0,sizeof(visit));        bfs();    }    return 0;}