PoJ3278--Catch That Cow(Bfs)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 61317		Accepted: 19155

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.

Source

USACO 2007 Open Silver

 

 

RE: 坐标轴上有两个点（a、b）, 求 a 接近 b 所需走的最小步数；

　　有三种走法，（x-1)、 （x+1）、（2*x）;

　　a > b  → → 输出；

　　else :: Bfs;

　　

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

const int maxn = 100000 + 5;
int vis[maxn], cnt[maxn];
queue <int> q;

void Bfs()
{
    int nx;
    while(!q.empty())
    {
        int x = q.front();  q.pop();
        for(int i = 0; i < 3; i++)    //三种走法； 
        {
            if(i == 0) nx = x + 1;
            else if(i == 1) nx = x - 1;
            else nx = 2 * x;
            if(nx >= 0 && nx <= 100000 && !vis[nx])  //遍历到最后找最优？::剪枝？ 
            {
                vis[nx] = 1;
                cnt[nx] = cnt[x] + 1;
                q.push(nx);     
            } 
        }     
    }
}

int main()
{
    int m, n;
    while(~scanf("%d %d", &m, &n))
    {
        if(m > n)
        {
            printf("%d\n", m - n);
            return 0;
        }
        memset(vis, 0, sizeof(vis));
        vis[m] = 1;
        cnt[m] = 0;
        q.push(m);
        Bfs();
        printf("%d\n", cnt[n]);
    } 
    return 0;
}