POJ 3278 Catch That Cow

题目地址： https://vjudge.net/problem/POJ-3278

 

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 129558		Accepted: 40251

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<cstring>   
#include<stdio.h>　
#include<queue>

using namespace std;

#define MAX 100000
int N, K;
int vis[MAX + 20];

struct node
{
    int x;
    int step;
}s;

void bfs(int x)
{
    memset(vis, 0, sizeof(vis));
    s.x = x;
    s.step = 0;
    vis[x] = 1;
    queue<node> Q;
    Q.push(s);

    node t;
    while (!Q.empty())
    {
        t = Q.front();
        Q.pop();

        if (t.x == K)
        {
            cout << t.step << endl;
            return;
        }

        if (t.x + 1 <= MAX && !vis[t.x + 1])
        {
            s.x = t.x + 1;
            s.step = t.step + 1;
            vis[s.x] = 1;
            Q.push(s);
        }
        if (t.x - 1 >= 0 && !vis[t.x - 1])
        {
            s.x = t.x - 1;
            s.step = t.step + 1;
            vis[s.x] = 1;
            Q.push(s);
        }
        if (t.x * 2 <= MAX && !vis[t.x * 2])
        {
            s.x = t.x * 2;
            s.step = t.step + 1;
            vis[s.x] = 1;
            Q.push(s);
        }
    }
    return;
}

int main()
{
    while(cin >> N >> K)
        bfs(N);


    return 0;
}