杭电 HDU 2717 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 8999    Accepted Submission(s): 2837

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.

睡前一水~ 好久没做过搜索了 注意剪枝&&&&&&&&
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
struct Node
{
    int x,step;
    Node(){}
    Node(int x,int step):x(x),step(step){}
};

int n,k;
bool vis[199999];

int bfs()
{
    queue<Node>Q;
     Node cur,next;
    Q.push(Node(n,0));
    while(!Q.empty())
    {
        cur=Q.front();
        Q.pop();
        for(int i=0;i<3;i++)
        {
            if(i==0)
                next.x=cur.x+1;
            if(i==1)
                next.x=cur.x-1;
            if(i==2)
                next.x=cur.x*2;
            next.step=cur.step+1;
            if(next.x==k)
                return next.step;
            if(next.x<0||next.x>100000)
                continue;
            if(!vis[next.x])
            {
                vis[next.x]=true;
                Q.push(next);
            }
        }
    }

}

int main()
{
    while(cin>>n>>k)
    {
        memset(vis,0,sizeof(vis));
        if(n<k)
            cout<<bfs()<<endl;
        if(n==k)
            cout<<0<<endl;
        if(n>k)
            cout<<n-k<<endl;
    }
    return 0;
}