POJ3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 68975 Accepted: 21699

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.

题意：初始点在N，目标点在K，但是每分钟只能向前或后退一步，或者到达这个点的2倍的点。最快要多久能从N点到达K点。

思路：  每次有三种情况，随意就是用了BFS，刚写成的时候，没有优化，导致超内存，后来修改，如果当前点超过了目标点K，就不再向前走，如果当前点小于0，就不能在后退，之后又TLE，然后发现没有优化那些走过的点，再次修改，终于AC了

代码如下：

#include<iostream>#include<cstdio>#include<queue>#include<cstring>#define MXAN 1000005using namespace std;struct Node{    int add,step;//add代表当前地址，step代表步数};int N,K;bool ok[MXAN];  //标记这点是否到达过queue<Node> Q;int bfs(){    int r,c;    Node tem={N,0};    Q.push(tem);    while(!Q.empty())    {        tem=Q.front();        if(tem.add==K) break;        Q.pop();        if(!ok[tem.add])   //如果这点遍历过，就不能在遍历        {            if(tem.add<K ) //如果这点小于目标点，可以继续+1或*2            {                r=tem.add+1;                Node temp={r,tem.step+1};                Q.push(temp);                c=tem.add*2;                Node tep={c,tem.step+1};                Q.push(tep);            }            if(tem.add>0 && !ok[tem.add])  //如果这点大于0，就能向后走            {                r=tem.add-1;                Node tp={r,tem.step+1};                Q.push(tp);            }            ok[tem.add]=true;  //把这点标记为已经遍历过        }    }    while(!Q.empty())    {        Q.pop();    }    return tem.step;}int main(){    memset(ok,0,sizeof(ok));    scanf("%d%d",&N,&K);    int ans=bfs();    printf("%d\n",ans);    return 0;}