Catch That Cow

Catch That Cow

Time Limit : 5000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 68   Accepted Submission(s) : 18

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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.

Source

USACO 2007 Open Silver

这题必须用广搜

广搜出来的是最小次

要注意 进队列 的数必须 没有扩充过 且 在0--100000内

#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
struct num
{
    int shu;//数目
    int chi;//扩充到第几次

};

int main()
{
    int star,end;
    while(scanf("%d %d",&star,&end)!=EOF)
    {
        num a1;
        a1.shu=star;//开始扩充的数
        a1.chi=0;
        bool used[100005];
        int number;//记录次数
        memset(used,0,sizeof(used));
        queue<num> q;
        q.push(a1);
        while(!q.empty())
        {
            num a2;
            a2=q.front();
            q.pop();
            if(a2.shu==end)//是目的数 退出
            {
                number=a2.chi;
                break;
            }
            if(used[a2.shu]==0)//没有被扩充过 进行扩充
            {
                used[a2.shu]=1;//标记
                num a3;
                a3.chi=a2.chi+1;
                a3.shu=a2.shu+1; //+1
                if(a3.shu>=0&&a3.shu<=100000&&used[a3.shu]==0)//没有被标记过且在范围内
                {
                    q.push(a3);
                }


                a3.chi=a2.chi+1;
                a3.shu=a2.shu-1; //-1

                if(a3.shu>=0&&a3.shu<=100000&&used[a3.shu]==0)//没有被标记过且在范围内
                {
                    q.push(a3);
                }


                a3.chi=a2.chi+1;
                a3.shu=a2.shu*2; //*2

                if(a3.shu>=0&&a3.shu<=100000&&used[a3.shu]==0)//没有被标记过且在范围内
                {q.push(a3);}

            }
        
        
        }
    
    printf("%d\n",number);
    }





}