| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 46459 | Accepted: 14574 |
Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX
+ 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
Output
Sample Input
5 17
Sample Output
4
Hint
Source
题目大意：农夫在n的位置，牛在k的位置，牛不动，若农夫任意时间在X，农夫可走到X+1，X-1,2*X三个位置，问农夫几次可抓到牛
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Queue;
import java.util.Scanner;
public class POJ3278_ieayoio {
public static boolean[] f;
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
while (cin.hasNext()) {
int n = cin.nextInt();
int k = cin.nextInt();
System.out.println(bfs(n, k));
}
}
static int bfs(int n, int k) {
// Queue<Node> queue = new LinkedList<Node>();
Queue<Node> queue = new ArrayDeque<Node>();
f = new boolean[100010];
Arrays.fill(f, true);//标记是否曾走过
Node node = new Node(n, 0);
f[n] = false;
queue.offer(node);//初始值直接入队
while (!queue.isEmpty()) {
node = queue.poll();//出队
if (node.location == k) {
return node.step;
}
Node el;
if (node.location < k && f[node.location + 1]) {
el = new Node(node.location + 1, node.step + 1);
f[node.location + 1] = false;
queue.offer(el);
}
if (node.location - 1 >= 0 && f[node.location - 1]) {
el = new Node(node.location - 1, node.step + 1);
f[node.location - 1] = false;
queue.offer(el);
}
//当node.location大于k是只能选择后退
//node.location * 2 <= 100000是坐标轴的范围，开始没加就Runtime Error了
if (node.location < k && node.location * 2 <= 100000
&& f[node.location * 2]) {
el = new Node(node.location * 2, node.step + 1);
f[node.location * 2] = false;
queue.offer(el);
}
}
return node.step;
}
}
class Node {
int location;//位置
int step;//步数
Node(int location, int step) {
this.location = location;
this.step = step;
}
}
POJ3278，Catch That Cow,布布扣,bubuko.com
原文地址：http://blog.csdn.net/ieayoio/article/details/38443627
