POJ 3278 Catch That Cow

POJ 3278 Catch That Cow

C - Catch That Cow

Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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

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.

題意：給你n,m兩個數，如何用最少操作次數把n變成m。n只能加一減一或者乘2。

范圍0到100000

直接bfs，越界或者搜過的數排除掉。

#include 
#define inf 0x6ffff
#include 
#include 
#include 
using namespace std;
struct node
{
	int x;
	int step;
}
;
int n,m;
int flag[100005];
int bfs(int x)
{
	int i;
	queueq;
	node st,ed;
	flag[x]=1;
	st.x=x;
	st.step=0;
	q.push(st);
	while(!q.empty())
	{
		st=q.front();
		q.pop();
		if(st.x==m )
			return st.step;
		for(i=1;i<=3;i++)
		{
			if(i==1)
				ed.x=st.x+1;
			if(i==2)
				ed.x=st.x-1;
			if(i==3)
				ed.x=st.x*2;	
			if(ed.x>100000 ||ed.x<0 ||flag[ed.x])  //搜過的數標記為一，越界排除。
				continue;
			ed.step=st.step+1;
			flag[ed.x]=1;
			q.push(ed);
		}
	}
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(flag,0,sizeof(flag));
		if(n==m)
			printf("0\n");
		else
		{
			int ans=bfs(n);
			printf("%d\n",ans);
		}
	}
	return 0; 
}