poj 3278

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 43851 Accepted: 13674

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

#include<iostream>#include<algorithm>#include<cstring>#include<queue>using namespace std;struct Node{    int x,t;    Node(int X,int T){        x=X; t=T;    }}; int sign[200010];   //注意这里的标记数组要定大一些，定为100010会runtime error int ans;queue <Node> q;int main(){    int n,k;    while(cin>>n>>k){        ans=9999999;        memset(sign,0,sizeof(sign));        while(!q.empty())            q.pop();   //清空队列         int x,t;        x=n; t=0;        q.push(Node(x,t));        //入队         sign[x]=1;                //标记已访问         while(!q.empty()){                    x=q.front().x; t=q.front().t;            q.pop();            if(x==k)                ans=min(ans,t);            else{                if(x<k){                    if(sign[x+1]==0){                        q.push(Node(x+1,t+1));                            sign[x+1]=1;                    }                    if(sign[2*x]==0){                        q.push(Node(2*x,t+1));                        sign[2*x]=1;                    }                    if(x>0&&sign[x-1]==0){                        q.push(Node(x-1,t+1));                        sign[x-1]=1;                    }                }                else if(x>k){                    if(sign[x-1]==0){                        q.push(Node(x-1,t+1));                        sign[x-1]=1;                     }                }            }        }         cout<<ans<<endl;     }    return 0;}