Problem 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.
思路:
向X-1、X+1和2*X进行bfs,开一个整型的dp数组,既进行了是否已经到达过的标记,又统计了步数。
上AC代码:
#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;
int n,k;
int dp[100001];
int main()
{
while(~scanf("%d%d",&n,&k))
{
memset(dp,-1,sizeof(dp));
queue<int> que;
que.push(n);
dp[n]=0;
while(!que.empty())
{
int now=que.front();
if(now==k)
{
break;
}
que.pop();
if(now-1>=0&&dp[now-1]==-1)
{
dp[now-1]=dp[now]+1;
que.push(now-1);
}
if(now+1<=100000&&dp[now+1]==-1)
{
dp[now+1]=dp[now]+1;
que.push(now+1);
}
if(2*now<=100000&&dp[2*now]==-1)
{
dp[2*now]=dp[now]+1;
que.push(2*now);
}
}
printf("%d\n",dp[k]);
}
return 0;
}