Catch That Cow

Time Limit: 2000MS	Memory Limit: 65536K
Total Submissions: 29250	Accepted: 9000

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.

Source

題目型別：搜尋 題目描述：略 題目分析：略 程式碼如下：

#include <stdio.h>
#include <string.h>
#define N 100001

int visit[N];
struct Point{
    int x;
    int step;
} queue[N*2],c,r,sp;

int bfs(int start,int target){
    int top = 0,end = 0;
    if(start == target){
        return 0;
    }
    memset(visit,0,sizeof(visit));
    visit[start] = 1;
    sp.x = start;
    sp.step = 0;
    queue[top] = sp;
    while(top <= end){
        c = queue[top];
        if(c.x + 1 < N && visit[c.x+1] == 0){
            visit[c.x+1] = 1;
            r.x = c.x+1;
            r.step = c.step+1;
            if(r.x == target){
                return r.step;
            }
            queue[++end] = r;
        }
        if(c.x - 1 >=0 && visit[c.x-1] == 0) {
            visit[c.x-1] = 1;
            r.x = c.x-1;
            r.step = c.step+1;
            if(r.x == target){
                return r.step;
            }
            queue[++end] = r;
        }
        if(c.x * 2 < N && visit[c.x*2] == 0){
            visit[c.x*2] = 1;
            r.x = c.x*2;
            r.step = c.step +1;
            if(r.x == target){
                return r.step;
            }
            queue[++end] = r;
        }
        top++;
    }
    return -1;
}

int main()
{
    int s,t;
    while(scanf("%d%d",&s,&t) != EOF){
        printf("%d\n",bfs(s,t));
    }

    return 0;
}