POJ 3278 Catch That Cow

解题思路:BFS

#include <iostream>
#include <cmath>
using namespace std;
#define MAXN 100001
int n, k, q[MAXN];
bool visit[MAXN];
int main()
{
    int s, e, p, i, j,step,minStep;
    while (scanf("%d %d", &n, &k)!=EOF)
    {
        q[0]=n,s=e=step=0,minStep=MAXN;
        if(n>=k)minStep=step=n-k;
        memset(visit, 0, sizeof(visit));
        while(s<=e&&step<minStep)
        {
            for(i=s,p=e;i<=e;i++)
            {
                if(q[i]==k&&minStep>step)minStep=step;
                if(q[i]-1>=0&&!visit[q[i]-1])q[++p]=q[i]-1,visit[q[p]]=true;
                if(q[i]+1<=k&&!visit[q[i]+1])q[++p]=q[i]+1, visit[q[p]]=true;
                if(2*q[i]<=k&&!visit[2*q[i]])q[++p]=2*q[i],visit[q[p]]=true;
                else if(2*q[i]>k)
                {
                    if(2*q[i]-k+step+1<minStep)minStep=2*q[i]-k+step+1;
                    if(k-q[i]+step<minStep)minStep=k-q[i]+step;
                }
            }
            step++,s=e+1,e=p;
        }
        printf("%d\n", minStep);
    }
    return 0;
}