ACM 之 B - 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.

Sample Input

5 17

Sample Output

4

[hit : (5->17) 5-10-9-18-17 , 一共走了4步 。]

理解

John的牛跑了，他要用最快的方式抓到这头牛，他可以选择后退一步或者前进一步甚至前进当前所在坐标的二倍，找出最快的路径帮他抓到牛。

代码部分

#include 
#include 
#include 
#include 
using namespace std;
int con,x,y,f,b[200000];//数组开到这里是极限，再小就会数据溢出。
queuea;
int main()
{
    while(scanf("%d%d",&x,&y)!=EOF)
    {
        while(!a.empty())
        {
            a.pop();
        }
        memset(b,0,sizeof(b));
        a.push(-2);
        con=0;
        f=0;
        if(x==y) 
            {printf("0\n");continue;}
        a.push(x);
        b[x]=1;
        int i=1;
        while(f!=1)
        {
            if(a.front()==-2)
            {
                a.pop();
                con++;
                a.push(-2);
            }
            if((a.front()-1)==y||(a.front()+1)==y||(a.front()*2)==y)
                {f=1;break;}
            if(b[a.front()-1]==0&&a.front()>1&&a.front()<100001)
                {a.push(a.front()-1);b[a.front()-1]=1;}//b数组是存在标记，表示出现过这个数了下次就不要再次运算，可以提高程序效率。
            if(b[a.front()+1]==0&&a.front()>-1&&a.front()<99999)
                {a.push(a.front()+1);b[a.front()+1]=1;}
            if(b[a.front()*2]==0&&a.front()>0&&a.front()<50001)
                {a.push(a.front()*2);b[a.front()*2]=1;}
            a.pop();
            //cout<

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