Catch That Cow（bfs）

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我是靠谱客的博主单身裙子，最近开发中收集的这篇文章主要介绍Catch That Cow（bfs），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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?

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

AC代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int N = 200010;
int s,g;
int visited[N];
struct Step
{
int x;//位置
int step;//步数
};
queue<Step> Q;
void bfs()
{
int X, STEP;
while(!Q.empty())
{
Step tmp = Q.front();
Q.pop();
X = tmp.x;
STEP = tmp.step;
if(X == g)
{
printf("%dn",STEP);
return ;
}
if(X>0 && !visited[X - 1]) //要保证减1后有意义,所以要X > 0
{
Step temp;
visited[X-1] = 1;
temp.x = X - 1;
temp.step = STEP + 1;
Q.push(temp);
}
if(X<=g && !visited[X + 1])
{
Step temp;
visited[X + 1] = 1;
temp.x = X + 1;
temp.step = STEP + 1;
Q.push(temp);
}
if(X<=g && !visited[X * 2])
{
Step temp;
visited[X * 2] = 1;
temp.x = 2*X;
temp.step = STEP + 1;
Q.push(temp);
}
}
}
int main ()
{
while(cin>>s>>g)
{
memset(visited,0,sizeof(visited));
Step tem;
tem.x=s;
tem.step=0;
Q.push(tem);
visited[s]=1;
bfs();
}
return 0;
}

主要思路借鉴慕课郭炜老师。