poj 3278 Catch That Cow(经典bfs)

题目：http://poj.org/problem?id=3278

Language: Default Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 62063   Accepted: 19415 Description Farmer John has 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 is at a 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 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 2 × 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:  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 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.

分析：将三种变化模拟成三条路，最终的结果是由A到达B的最短时间，所以联想到BFS。
虽然很多人说这是一道水题，但我却在前几次都WA了(555 T_T)  哎，走过的点就不该再走了！标记一下。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1e5+10;
typedef long long LL;
LL n,k;
struct node{
    LL dex,step;
    LL go(LL x){
        if(x==1) return dex+1;
        else if(x==2) return dex-1;
        else if(x==3) return dex*2;
    }
}p[maxn];
bool vis[maxn];
node que[maxn];
void bfs(node start){
    LL q1=0,q2=0;
    que[q2++]=start;
    vis[start.dex]++;
    while(q1!=q2){
         node cur=que[q1];
         q1=(q1+1)%(maxn-9);
         for(LL i=1;i<=3;i++){
             LL dex=cur.go(i);
             if(dex>maxn-10 || dex<0 || vis[dex]>0) continue;
             p[dex].step=cur.step+1;
             que[q2]=p[dex];
             q2=(q2+1)%(maxn-9);
             vis[dex]++;
             if(dex==k) return ;
         }
    }
}
int main()
{
    //freopen("cin.txt","r",stdin);
    while(cin>>n>>k){
        if(n==k) { puts("0");  continue;  }
        memset(vis,0,sizeof(vis));
        for(LL i=0;i<maxn;i++){
            p[i].dex=i;
            p[i].step=0;
        }
        bfs(p[n]);
        printf("%lld\n",p[k].step);
    }
    return 0;
}