【BFS】【HDOJ-2717】Catch That Cow

Problem 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

 

思路：

当n>=k的时候只能后退一步 所以就是n-k

n<k的时候有三种 +1 -1 *2 BFS

 

参考代码

#include <stdio.h>
#include <string.h>
int bfs(int n,int k);
#define LEN 100001
int step[LEN]={0};
int mark[LEN]={0};
int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n>=k)
            printf("%d
",n-k);
        else
            printf("%d
",bfs(n,k));
    }
    return 0;
}

int bfs(int n,int k)
{
    memset(step,0,sizeof(step));
    memset(mark,0,sizeof(mark));
    int queue[LEN];
    int i,front=0,rear=1;
    queue[0]=n;
    mark[queue[0]]=1;
    while(front<rear)
    {
        for(i=0;i<3;i++)
        {
            if(queue[front]-1>=0&&!mark[queue[front]-1])
            {
                mark[queue[front]-1]=1;
                step[queue[front]-1]=step[queue[front]]+1;
                queue[rear++]=queue[front]-1;
                if(queue[rear-1]==k)
                    return  step[queue[front]-1];
            }
            if(queue[front]+1<=LEN&&!mark[queue[front]+1])
            {
                mark[queue[front]+1]=1;
                step[queue[front]+1]=step[queue[front]]+1;
                queue[rear++]=queue[front]+1;
                if(queue[rear-1]==k)
                    return step[queue[front]+1];
            }
            if(queue[front]*2<=LEN&&!mark[queue[front]*2])
            {
                mark[queue[front]*2]=1;
                step[queue[front]*2]=step[queue[front]]+1;
                queue[rear++]=2*queue[front];
                if(queue[rear-1]==k)
                    return step[queue[front]*2];
            }
        }
        front++;
    }
}