Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 80273 | Accepted: 25290 |
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
题意：给定起点和终点，有三种走的方式，假设当前为 now，那么可以选择下一次到达 now - 1 或 now + 1 或 2*now，问起点到终点最少需要几步。
题解：bfs，下一个状态即为三种可选择的位置。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 100005;
int step[maxn],que[maxn],vis[maxn];
int bfs(int st,int ed)
{
int E = 0,F = 0;
que[F++] = st;
vis[st] = 1;
for (;;)
{
int now = que[E];
if (now == ed) return step[now];
if (now + 1 < maxn && !vis[now + 1]) step[now + 1] = step[now] + 1,vis[now + 1] = 1,que[F++] = now + 1;
if (now - 1 >= 0 &&!vis[now - 1]) step[now - 1] = step[now] + 1,vis[now - 1] = 1,que[F++] = now - 1;
if (2*now < maxn && !vis[2*now]) step[2*now] = step[now] + 1,vis[2*now] = 1,que[F++] = 2 * now;
E++;
}
}
int main()
{
int N,K;
memset(vis,0,sizeof(vis));
scanf("%d%d",&N,&K);
if (N >= K) printf("%d\n", N - K);
else printf("%d\n",bfs(N,K));
return 0;
}