暑假集训第三周 STL J - Catch That Cow 抓捕奶牛

J - Catch That Cow

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

Submit  Status

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.

分析：

这是一个广搜+队列的问题，很复杂的问题用这些方法就能很轻松的写出来。

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#include<iostream>
#include<queue>
#include<stdio.h>
#include<string.h>
using namespace std;
queue<int> q;
int step[100001],visit[100001];
int  main()
{
    int a,b;
    while(scanf("%d %d",&a,&b)!=EOF)
    {
        memset(visit,0,sizeof(visit));
        if(a>=b)
            printf("%d\n",a-b);
        else
        {
            int t,temp;
            queue<int>q;
            q.push(a);  // 插入队列
            visit[a]=1;
            while(!q.empty())
            {
                t=q.front();//队列首元素
                q.pop();   //出队列
                for(int i=0; i<3; i++)
                {
                    if(i==0)
                        temp=t+1;
                    else if(i==1)
                        temp=t-1;
                    else
                        temp=t*2;

                    if(temp>100000||temp<0)
                        continue;
                    if(visit[temp]==0)
                    {
                        step[temp]=step[t]+1;
                        if(temp==b)
                            printf("%d\n",step[temp]);
                        visit[temp]=1;
                        q.push(temp);

                    }
                }
            }
        }
    }
    return 0;
}