POJ 3278 Catch that Cow (BFS) __ Search

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her. He starts in a point N (0≤n≤100,000) on a number line and the cow are at 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 points X-1 or X + 1 in a single minute

TELEPORTING:FJ can move from all point X to the "2xX in a" single minute.

If the cow, unaware of it pursuit, does not move at all, how long is 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

the

The farmer needs to find his cattle, the farmer is now in the position of N, the cow at K Point, assuming the cow is not moving, the farmer has three kinds of walk: step forward (n+1) step Backward (N-1) jump to the current position twice times (2*n)

Ask, how many times to walk to find his cattle.

train of Thought

Simple BFS, the farmer's current position into the queue, and then out of the team to judge each point, and simulate the next move, Note that the need to mark the point of passing, or will burst stack.

AC Code

#include <iostream> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <

queue> #include <math.h> #define MAXX 100005 using namespace std;
BOOL Isvisted[maxx];
    struct node {int n;
    int time;
        Node (int n,int time) {this->n=n;
    this->time=time;
}
};
    void BFs (int n,int k) {memset (isvisted,false,sizeof (isvisted));
    Node Ans=node (n,0);
    Isvisted[n]=true;
    queue<node>sk;
    Sk.push (ANS);
        while (!sk.empty ()) {node P=sk.front ();
        Sk.pop ();     Isvisted[p.n]=true;
            Mark traversed by the dot if (p.n==k)///find {ans=p;
        Break
        } if (p.n+1<maxx&&!isvisted[p.n+1])//Three kinds of Sk.push (node (p.n+1,p.time+1));
        if (p.n-1>=0&&!isvisted[p.n-1]) Sk.push (node (p.n-1,p.time+1));
    if (p.n*2<maxx&&!isvisted[p.n*2]) Sk.push (node (p.n*2,p.time+1)); } priNTF ("%d\n", ans.time);
    int main () {int n,k;
    while (~SCANF ("%d%d", &n,&k)) BFS (n,k);
return 0; }