Problem C POJ 3278 Catch That Cow(三入口bfs)

C - Catch That Cow

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3278

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.

 

题目大意：

　　给你一个数字n,n<=100000，然后，再给你一个数字k，求把k变成n所需要的最小的步数.

解题思路：

　　看到最小步数，想都不都想直接bfs，因为bfs是能够把我们在第N步的N个状态全部扩展出来，所以找起来更快

　　一开始wa了一发，原因是对于n==k的情况，答案应该是0的，但是我的程序跑出来却是2，原因就是把bfs写惨了

　　每次在进start的时候，判断下就好了、

代码：

# include<cstdio>
# include<iostream>
# include<cstring>
# include<queue>

using namespace std;

# define MAX 100000+4

struct node
{
    int x;
    int step;
};
queue<node>Q;

int book[MAX];
int n,k;

void init()
{
    while ( !Q.empty() )
    {
        Q.pop();
    }
    memset(book,0,sizeof(book));
}

int bfs ( node start )
{
    init();
    Q.push(start);
    if ( start.x==k )
        return start.step;
    while (!Q.empty())
    {
        node now = Q.front();
        Q.pop();
        int tx = 0;
        for ( int i = 0;i < 3;i++ )
        {
            if ( i == 0 )
                tx = now.x-1;
            else if ( i==1 )
                tx = now.x+1;
            else
                tx = 2*now.x;
            if ( tx > MAX||book[tx]==1 )
                continue;
            if ( tx==k )
                return now.step+1;
            node newnode;
            if ( book[tx]==0 )
            {
                book[tx] = 1;
                newnode.x = tx; newnode.step = now.step+1;
                Q.push(newnode);
            }
        }
    }
}


int  main(void)
{
    while ( scanf("%d%d",&n,&k)!=EOF )
    {
        book[n] = 1;
        node start;
        start.step = 0; start.x = n;
        int ans = bfs(start);
        printf("%d
",ans);
    }

    return 0;
}