C - 青铜五 HDU - 2717 Catch That Cow BFS

C - 青铜五

 HDU - 2717 

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：一维地图上有两个点，最少多少步可以 从起点到终点；人可以：1.向左或右走一步；2.直接从当前位置x传送到2*x

搜索，没啥好说的

扫描二维码关注公众号，回复： 2566247 查看本文章

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <math.h>

typedef long long LL;
typedef long double LD;
using namespace std;
const int  maxn=100000*2;
int vis[maxn];
int f[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
int n,m;
struct node
{
    int x;
    int step;
};
queue<node>q;

void bfs(node st,node en)
{
    q.push(st);
    //printf("*****\n");
    while(!q.empty())
    {
        node t=q.front();
        //printf("%d %d\n",t.x,t.step);
        q.pop();
        if(t.x==en.x)
        {
            printf("%d\n",t.step);
            return;
        }
        if(t.x+1<=en.x&&vis[t.x+1]==0)
        {
            q.push((node){t.x+1,t.step+1});
            vis[t.x+1]=1;
        }
        if(t.x-1>=0&&vis[t.x-1]==0)
        {
            q.push((node){t.x-1,t.step+1});
            vis[t.x-1]=1;
        }
        if(t.x*2<=2*en.x&&vis[t.x*2]==0)
        {
            q.push((node){t.x*2,t.step+1});
            vis[t.x*2]=1;
        }
    }
}
int main()
{
    int N,K;
    while(~scanf("%d%d",&N,&K))
    {
        //printf("$$$$$$$$$$$$\n");
        node st,en;
        st.x=N;
        st.step=0;
        en.x=K;
        memset(vis,0,sizeof(vis));
        vis[N]=1;
        while(!q.empty())q.pop();
        bfs(st,en);

    }
    return 0;
}