Poj 3278 catch that cow (broad search)

Catch that cow Time limit:2000 ms   Memory limit:65536 K Total submissions:45087   Accepted:14116 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 pointK(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 pointXTo the pointsX-1 orX+ 1 in a single minute * Teleporting: FJ can move from any pointXTo the point 2 ×XIn 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:NAndK 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. Source Usaco 2007 Open silver

The meaning of the question tells you to calculate the movement of the shortest steps from N to K at two points n k to n + 1 or n-1 or N * 2;

Search for the shortest path in three directions

#include<iostream>#include<cstdio>#include<cstring>const int M=200005;using namespace std;typedef class{    public:    int x;    int step;}wz;wz q[M];int vis[M];int n,k;int bfs(){    int front=0,rear=0;    q[rear].x=n;    q[front].x=n;    q[rear++].step=0;    vis[n]=1;    while(front<rear)    {       wz w=q[front++];        if(w.x==k)            return w.step;        if(w.x-1>=0&&!vis[w.x-1])        {            vis[w.x-1]=1;            q[rear].x=w.x-1;            q[rear++].step=w.step+1;        }        if(w.x+1<=k&&!vis[w.x+1])        {            vis[w.x+1]=1;            q[rear].x=w.x+1;            q[rear++].step=w.step+1;        }        if(w.x<=k&&!vis[2*w.x])        {            vis[2*w.x]=1;            q[rear].x=w.x*2;            q[rear++].step=w.step+1;        }    }}int main(){    int i;    while(cin>>n>>k)    {        memset(vis,0,sizeof vis);          cout<<bfs()<<endl;    }    return 0;}