POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 43674		Accepted: 13612

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

题目大意：

告诉你起点位置和终点位置，你可以有三种操作，从某个位置x，可以到达x-1,x+1,2*x，问你从起点到终点最短时间？

解题思路：

利用bfs枚举位置即可

但是刨除以下情况，也就是剪枝一下

1、当x<=0 时，x-1不能到达，这个状态剪掉

2、当x>终点位置时，x+1和2*x均不能到达

解题代码：

#include <iostream>
#include <queue>
using namespace std;

const int maxn=210000;
int marked,visited[maxn];

struct node{
    int pos,t;
    node(int pos0=0,int t0=0){
        pos=pos0;t=t0;
    }
};

int main(){
    int from,to;
    while(cin>>from>>to){
        marked++;
        queue <node> q;
        q.push(node(from,0));
        visited[from]=marked;
        while(!q.empty()){
            node s=q.front();
            q.pop();
            if(s.pos==to){
                cout<<s.t<<endl;
                break;
            }
            if( s.pos-1>=0 && visited[s.pos-1]!=marked){
                visited[s.pos-1]=marked;
                q.push(node(s.pos-1,s.t+1));
            }
            if( s.pos<=to && visited[s.pos+1]!=marked){
                visited[s.pos+1]=marked;
                q.push(node(s.pos+1,s.t+1));
            }
            if( s.pos<=to && visited[s.pos*2]!=marked){
                visited[s.pos*2]=marked;
                q.push(node(s.pos*2,s.t+1));
            }
        }
    }
    return 0;
}