Catch 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

提示

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.

示例程序

#include <stdio.h>#include <stdlib.h>#include <string.h>int n, k;int que[100010];int book[100010];int has(int i){    if(i<0 || i>100000||book[i]){        return 0;    }else {        return 1;    }}int bfs(int n,int k){    if(n == k) return 0;    int head = 0;    int tail = 1;    que[head] = n;    int x;    while(head < tail){        x = que[head];        head++;        if(x+1==k || x-1==k || x*2 == k){            return book[x]+1;        }        if(has(x-1)){            book[x-1] = book[x]+1;            que[tail++] = x-1;        }        if(has(x+1)){            book[x+1] = book[x]+1;            que[tail++] = x+1;        }        if(has(x*2)){            book[x*2] = book[x]+1;            que[tail++] = x*2;        }    }    return -1;}int main(){    while(~scanf("%d %d", &n, &k)){        memset(book,0,sizeof(book));        memset(que,0,sizeof(que));        printf("%d\n", bfs(n,k));    }    return 0;}