POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 51447		Accepted: 16134

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：

     给出起始位置n，有三种操作x+1,x-1,2*x,求最少的操作次数，使x==k。

题解：

     广搜，裸的广搜会MLE，要加剪枝，还要用数组标记。一直MLE，就调整为每次出现一个新位置就判断是否结束，导致后来忘记考虑n==k的情况%>_<%

    剪枝1.x>k,2*x不用走，只会越走越远。

    剪枝2.x>k,x+1不用走，使x变小的只有x-1，x+1再x-1只会增加步数。

    剪枝3.x-1>=0,x为负数时，只会增加步数或越走越远。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
int n,k;
int ans;
const int maxn=200000+100;
struct node
{
    int x;
    int cur;
} p;
bool vis[maxn];
queue<node> q;
void BFS()
{
    node p;
    p.x=n;
    p.cur=0;
    vis[n]=true;
    while(!q.empty())
        q.pop();
    q.push(p);
    while(!q.empty())
    {
        node now;
        node tmp=q.front();
        q.pop();
        if(tmp.x==k)//n==k时
        {
            ans=tmp.cur;
            return;
        }
        if(tmp.x<k&&!vis[tmp.x*2])//剪枝
        {
            now.x=tmp.x*2;
            now.cur=tmp.cur+1;
            vis[now.x]=true;
            if(now.x==k)
            {
                ans=now.cur;
                return;
            }
            q.push(now);
        }
        if(tmp.x<k&&!vis[tmp.x+1])//剪枝
        {
            now.x=tmp.x+1;
            now.cur=tmp.cur+1;
            vis[now.x]=true;
            if(now.x==k)
            {
                ans=now.cur;
                return;
            }
            q.push(now);
        }
        if(tmp.x-1>=0&&!vis[tmp.x-1])//剪枝
        {
            now.x=tmp.x-1;
            now.cur=tmp.cur+1;
            vis[now.x]=true;
            if(now.x==k)
            {
                ans=now.cur;
                return;
            }
            q.push(now);
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        memset(vis,false,sizeof(vis));
        BFS();
        printf("%d\n",ans);
    }
    return 0;
}