[HDOJ2717]Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8616    Accepted Submission(s): 2714

Problem 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.

 

 

经典BFS，常规思路。

注意：

题目所给5 17和17 5答案是不一样的。我曾天真地认为需要进行一步swap，实际上是不需要的。

遍历到某点应判断是否越界，否则也会ACCESS_VIOLATION。

数组也要开大一点，否则也会ACCESS_VIOLATION。

（PS：我忘记修改判断是否越界时界限的大小。）

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int vis[200010];
int stp[200010];
int n, k;

void swap(int& a, int& b)
{
    a = a ^ b;
    b = a ^ b;
    a = a ^ b;
}

/*
int move(int sgn, int cur)
{
    if(sgn == 1)
    {
        return cur + 1;
    }
    if(sgn == -1)
    {
        return cur - 1;
    }
    else
    {
        return cur * 2;
    }
}
*/

void BFS(int ini)
{
    int cur, now = 0;    //init
    queue<int> q;
    vis[ini] = 1;
    stp[ini] = 0;
    q.push(ini);
    while(!q.empty())
    {
        cur = q.front();
        q.pop();
        for(int i = 0; i < 3; i++)
        {
            if(i == 0)
            {
                now = cur + 1;
            }
            else if(i == 1)
            {
                now = cur - 1;
            }
            else
            {
                now = cur * 2;
            }

            if(!vis[now] && now >= 0 && now <= 100010)
            {
                vis[now] = 1;
                q.push(now);
                stp[now] = stp[cur] + 1;
            }
            if(now == k)
            {
                printf("%d\n", stp[now]);
                return ;
            }
        }
    }
}
int main()
{
    while(scanf("%d %d", &n, &k) != EOF && n+k)
    {
        memset(vis, 0, sizeof(vis));
        memset(stp, 0, sizeof(stp));
        if (n == k)
        {
            printf("0\n");
        }
        else
        {
            BFS(n);
        }
    }
    return 0;
}