bzoj1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

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?

    农夫约翰被通知,他的一只奶牛逃逸了!所以他决定,马上幽发,尽快把那只奶牛抓回来.
    他们都站在数轴上.约翰在N(O≤N≤100000)处,奶牛在K(O≤K≤100000)处.约翰有
两种办法移动,步行和瞬移:步行每秒种可以让约翰从z处走到x+l或x-l处;而瞬移则可让他在1秒内从x处消失,在2x处出现.然而那只逃逸的奶牛,悲剧地没有发现自己的处境多么糟糕,正站在那儿一动不动.
    那么,约翰需要多少时间抓住那只牛呢?

Input

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

    仅有两个整数N和K.

 

Output

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

    最短的时间.

Sample Input

5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.

Sample Output

4

OUTPUT DETAILS:

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.
 

 
当$n<=1e5$时,它就是道简单的bfs
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<queue>
 5 using namespace std;
 6 #define N 100002
 7 int vis[N<<1],s,t;
 8 queue <int> h;
 9 int main(){
10     scanf("%d%d",&s,&t); h.push(s);
11     if(s==t){printf("0");return 0;}
12     while(!h.empty()){
13         int x=h.front(); h.pop();
14         if(!vis[x-1]&&x) vis[x-1]=vis[x]+1,h.push(x-1);
15         if(!vis[x+1]&&x+1<N) vis[x+1]=vis[x]+1,h.push(x+1);
16         if(!vis[x<<1]&&(x<<1)<N) vis[x<<1]=vis[x]+1,h.push(x<<1);
17         if(vis[t]){printf("%d",vis[t]);return 0;}
18     }
19 }
View Code

 

 
posted @ 2018-11-28 14:00  kafuuchino  阅读(107)  评论(0编辑  收藏  举报