POJ(3278)Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 74588		Accepted: 23516

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.

一个人追一头牛，给出人的位置，与走法，问人能多少秒追到牛：

简单搜索bfs

#include <iostream>
#include<algorithm>
#include<stdio.h>
#include <string.h>
#include <queue>
using namespace std;
int h[1000010],v[1000010],n,k;
void bfs()
{
memset(v,0,sizeof(v));
queue<int>q;
q.push(n);
v
=1;
h
=0;
while(!q.empty())
{
int temp=q.front();
q.pop();
if(temp+1==k||temp-1==k||temp*2==k)
{
h[k]=h[temp]+1;
break;
}
if(temp-1>=0&&!v[temp-1])
{
q.push(temp-1);
v[temp-1]=1;
h[temp-1]=h[temp]+1;
}
if(temp+1<=1000000&&!v[temp+1])
{
q.push(temp+1);
v[temp+1]=1;
h[temp+1]=h[temp]+1;
}
if(temp*2<=1000000&&!v[temp*2])
{
q.push(temp*2);
v[temp*2]=1;
h[temp*2]=h[temp]+1;
}
}
}

int main()
{
while(~scanf("%d%d",&n,&k))
{
if(n==k)
{
printf("0\n");
continue;
}
else
bfs();
printf("%d\n",h[k]);
}
return 0;
}