SDUT1028Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory limit: 65536K

题目描述

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?

输入

Line 1: Two space-separated integers: N and K

输出

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

示例输入

5 17

示例输出

4

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct node
{
    int x;
    int ans;
}q[1001000];
int n,m;
int jilu[1001000];
void bfs(int x,int ans)
{
    struct node t,f;
    f.x=x;
    f.ans=0;
    jilu[f.x]=1;
    int jin=0,chu=0;
    q[jin++]=f;
    int i;
    while(jin>chu)
    {
        t=q[chu++];
        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&&jilu[f.x]!=1)
               {
                   f.ans=t.ans+1;
                   jilu[f.x]=1;
                   q[jin++]=f;
               }

        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(jilu,0,sizeof(jilu));
        bfs(n,0);
    }
    return 0;
}