poj3278Catch That Cow

最新推荐文章于 2024-09-04 11:33:10 发布

原创最新推荐文章于 2024-09-04 11:33:10 发布 · 254 阅读

0 ·

CC 4.0 BY-SA版权

大学刷题专栏收录该内容

276 篇文章

订阅专栏

本文介绍了一个有趣的算法问题——如何让农夫约翰用最短的时间抓到一只静止不动的逃逸奶牛。通过使用广度优先搜索算法，文章提供了一种解决此问题的有效方法，并附带实现代码。

Catch That Cow

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status

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

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.

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
const int man=1e7+5;
using namespace std;
int visit[man];
int n,k;
struct node{
int x,num;
};
bool operator< (const node &a,const node &b)
{
return a.num>b.num;
}
void bfs()
{
priority_queue<node>q;
node star,end;
star.x=n;
star.num=0;
q.push(star);
memset(visit,0,sizeof(visit));
visit[n]=1;
while(!q.empty())
{
star=q.top();
q.pop();
if(star.x==k)
{
printf("%d\n",star.num);
break;
}
int move[3]={1,-1,star.x};
for(int i=0;i<3;i++)
{
end.x=star.x+move[i];
if(!visit[end.x]&&(end.x>=0)&&(end.x<=100000))
{
visit[end.x]=1;
end.num=star.num+1;
q.push(end);
}
}
}
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
bfs();
}
return 0;
}