POJ 3278 Catch that Cow

Catch that Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 71439		Accepted: 22496

Description

Farmer John had 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 are at a point K (0≤ K ≤100,000) on the same number line. Farmer John has 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 2x 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: Nand 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-Farmer John to reach the fugitive cow are to move along the following PATH:5-10-9-18-17, which takes 4 Minutes.

Re several times, the max value is changed to more than 200,000 AC. However, the running time is more than 2000 MS, it feels Java is really much slower than C + +.

Java AC Code

Importjava.util.LinkedList;ImportJava.util.Queue;ImportJava.util.Scanner; Public classMain {Staticqueue<integer> queue =NewLinkedlist<integer>(); Static intmax = 200010; Static BooleanMarked[] =New Boolean[Max]; Static intSteps[] =New int[Max]; Static intBoy , Cow;  Public Static voidMain (string[] args) {Scanner sc=NewScanner (system.in); Boy=Sc.nextint (); Cow=Sc.nextint ();        BFS ();    System.out.println (Steps[cow]); }         Public Static voidBFs () {queue.add (boy); Steps[boy]= 0; Marked[boy]=true;  while(!Queue.isempty ()) {            intHead =Queue.poll (); if(head-1 >= 0 &&!marked[head-1]) {Steps[head-1] = Steps[head] + 1; Marked[head-1] =true; Queue.add (Head+ W); }            if(head + 1 < max &&!marked[head + 1]) {Steps[head+ 1] = Steps[head] + 1; Marked[head+ 1] =true; Queue.add (Head+ 1); }            if(Head * 2 < Max &&!marked[head * 2]) {Steps[head* 2] = Steps[head] + 1; Marked[head* 2] =true; Queue.add (Head* 2); }            if(Marked[cow])return; }    }    }

POJ 3278 Catch that Cow