POJ3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total 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 1000005
using 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;
}