poj3278

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 71110 Accepted: 22371

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

USACO 2007 Open Silver

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<malloc.h>#include<math.h>struct node{    int step;    int loc;};struct node que[1000005];int book[400005];int main(){    int start,end;    int head=0;    int tail=0;    scanf("%d %d",&start,&end);    if(start>=end)        printf("%d\n",start-end);    else{        que[tail].loc=start;        que[tail].step=0;        book[start]=1;        tail++;        int i,j;        int t=0;        int flag=0;        while(head<tail){            for(i=1;i<=3;i++){                if(i==1){                    t=que[head].loc-1;                }                else if(i==2){                    t=que[head].loc+1;                }                else if(i==3){                    t=que[head].loc*2;                }                if(t<0||t>100005)                    continue;                if(book[t]==0){                    book[t]=1;                    que[tail].loc=t;                    que[tail].step=que[head].step+1;                    tail++;                }                if(t==end){                    flag=1;                    break;                }            }            if(flag)                break;            head++;        }        printf("%d\n",que[tail-1].step);    }    return 0;}

注意考虑起点和终点一样的情况