poj 3278 广搜

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 70333		Accepted: 22115

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.

题意：就是通过三种方式从一个点到另一个点

这里使用队列可以减少空间的使用，用数组模拟就不知道要多大的空间了

注意这里搜索 5 到时 17 是用了4 步，

但是17到 5  就会用 12 步，注意题目说的运动顺序

a[m-1]=a[m]+1;

上面这一步意思是：用了多少步走到点 m-1。其他同理可以得到

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;


const int maxn=100001;
int used[maxn];
int a[maxn];


int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        memset(used,0,sizeof(used));
        memset(a,0,sizeof(a));

        queue<int>q;
        q.push(m);
        used[m]=1;
        while(q.size()!=0)
        {
            m=q.front();
            q.pop();


            if(m==n)
                break;
            if(m-1>=0&&used[m-1]==0){
                q.push(m-1);
                used[m-1]=1;
                a[m-1]=a[m]+1;
            }
            if(m+1<=maxn&&used[m+1]==0){
                q.push(m+1);
                used[m+1]=1;
                a[m+1]=a[m]+1;
            }
            if(m*2<=maxn&&used[m*2]==0){
                q.push(m*2);
                used[m*2]=1;
                a[m*2]=a[m]+1;
            }
        }
        printf("%d\n",a[n]);
    }
    return 0;
}