poj 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 29149		Accepted: 8967

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.

 

#include<iostream>
#include<queue>
using namespace std;

int step[100001],map[100001];

int judge(int a)
{
    if(a<0 || a>100000 || map[a] == 1)
        return 0;
    return 1;
}

int main()
{
    int k, n, y, m;
    queue<int>x;
    memset(map, 0, sizeof(map));
    memset(step, 0, sizeof(step));

    cin>>n>>k;
    x.push(n);
    step[n] = 0;
    map[n] = 1;
    while(!x.empty())
    {
        y=x.front();
        x.pop();
        map[y]=1;

        if(y==k)
            break;

        m = y-1;
        if(judge(m))
        {
            x.push(m);
            step[m]=step[y]+1;
            map[m]=1;
        }
        m = y+1;
        if(judge(m))
        {
            x.push(m);
            step[m]=step[y]+1;
            map[m]=1;
        }
        m = y*2;
        if(judge(m))
        {
            x.push(m);
            step[m]=step[y]+1;
            map[m]=1;
        }
    }
    cout<<step[y]<<endl;
    return 0;
}

一维的BFS，更简单了