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;}