Catch That Cow oj

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KB

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.

Example Input

5 17

Example Output

4

Hint

poj3278 有链接提示的题目请先去链接处提交程序，AC后提交到SDUTOJ中，以便查询存档。
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.

Author

#include <stdio.h>
struct node
{
    int x;
    int ans;
}q[1001000];
int n,m;
int map[1001000];
void bfs(int x,int ans)
{
    struct node t,f;
    f.x=x;
    f.ans=0;
    map[f.x]=1;
    int head=0,last=0;
    q[head++]=f;
    int i;
    while(head>last)
    {
        t=q[last++];
        if(t.x==m)
        {
            printf("%d\n",t.ans);
            break;
        }
        for(i=0;i<3;i++)
        {
            if(i==0)
                f.x=t.x+1;
            if(i==1)
                f.x=t.x-1;
            if(i==2)
                f.x=t.x*2;
            if(f.x<=100000&&map[f.x]==0)
            {
                f.ans=t.ans+1;
                map[f.x]=1;
                q[head++]=f;
            }
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&m) != EOF)
    {
        memset(map,0,sizeof(map));
        bfs(n,0);
    }
    return 0;
}