Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 89836 | Accepted: 28175 |
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.
起点在n，终点是k，每一步可以是+1、-1或*2，最少走几步？
#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
int a[];//一个2倍大小的数组代表可以走到的位置，因为有乘二所以要开二倍以防RE，a[i]=x --> 走到坐标i需要x步
int main()
{
int n, k, t;
queue<int>q;
scanf("%d%d", &n, &k);
memset(a, -, sizeof(a));//所有坐标初始化为-1
a[n] = ; //n走到n当然是需要步
q.push(n); //当前位点入队
while (!q.empty())
{
t = q.front(); //读取队首
q.pop(); //删除队首
if (t == k) //若到达终点则直接输出并结束
{
printf("%d\n", a[k]);
return ;
}
if (t - >= && a[t - ] == -)//可能到达的结点入队，要判断是否越界，走过的不再走
{
a[t - ] = a[t] + ; q.push(t - );//下一步走到的位点所需步数是当前位点的步数+1
}
if (t + < && a[t + ] == -)
{
q.push(t + ); a[t + ] = a[t] + ;
}
if (t * < && a[t * ] == -)
{
q.push(t * ); a[t * ] = a[t] + ;
}
}
}
简单的bfs