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: 65536KTotal 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;}