[VJ][bfs]Catch That Cow

Catch That Cow

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.

Examples

Input

5 17

Output

4

描述：

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.

正确解法：

以前的都是二维，突然一维有点不会qaq

不能超范围，n==k时要特殊判断。

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int n, k;
int book[100010];
struct student
{
    int x, step;
}que[100010];
bool  check(int x)
{
    if (x < 0 || x>100000 || book[x] == 1)
        return 0;
    return 1;
}
int bfs()
{
    if (n == k)    return 0;
    book[n] = 1;
    int head = 1, tail = 2;
    que[head].x = n;
    que[head].step = 0;
    while (head < tail) {
        int tx1 = que[head].x - 1;
        if (check(tx1))
        {
            que[tail].x = tx1;
            que[tail].step = que[head].step + 1;
            book[tx1] = 1;
            if (tx1 == k)
                return que[tail].step;
            tail++;
        }
        int tx2 = que[head].x + 1;
        if (check(tx2))
        {
            que[tail].x = tx2;
            que[tail].step = que[head].step + 1;
            book[tx2] = 1;
            if (tx2 == k)
                return que[tail].step;
            tail++;
        }
        int tx3 = que[head].x * 2;
        if (check(tx3))
        {
            que[tail].x = tx3;
            que[tail].step = que[head].step + 1;
            book[tx3] = 1;
            if (tx3 == k)
                return que[tail].step;
            tail++;
        }
        head++;
    }
}
int main()
{
    cin >> n >> k;
    cout << bfs() << endl;
    return 0;
}