POJ 3278 Catch That Cow（bfs）

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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。

 

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#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; }