Catch That Cow（poj 3278）

给定两个整数n和k

通过 n+1或n-1 或n*2 这3种操作，使得n==k

输出最少的操作次数

//广搜，a是队列，step记录步数，vis记录哪些数被搜到过
#include<cstdio>
#include<iostream>
#define M 1000010
using namespace std;
int step[M],a[M],vis[M];
int main()
{
    int n,m,head=,tail=;
    scanf("%d%d",&n,&m);
    a[]=n;
    vis[n]=;
    while(head<tail)
    {
        ++head;
        int u=a[head];
        if(u==m)
        {
            printf("%d",step[head]);
            return ;
        }
        if(u->=&&!vis[u-])
        {
            a[++tail]=u-;
            step[tail]=step[head]+;
            vis[u-]=;
        }
        if(u+<&&!vis[u+])
        {
            a[++tail]=u+;
            step[tail]=step[head]+;
            vis[u+]=;
        }
        if(u*<&&!vis[u*])
        {
            a[++tail]=u*;
            step[tail]=step[head]+;
            vis[u*]=;
        }
    }
    return ;
}