Catch That Cow

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概述

Catch That Cow

Time Limit:2000MS Memory Limit:65536K
Total Submit:118 Accepted:30

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

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

#include<iostream> #include<queue> using namespace std; #define M 100001 #define INF 10000000 struct POINT { int pos; int step; }now,next; queue<POINT>Q; bool visited[M]; int n,k; int bfs() { while(!Q.empty()) Q.pop(); now.pos=n; now.step=0; visited[now.pos]=true; Q.push(now); while(!Q.empty()) { now=Q.front(); Q.pop(); next=now; for(int i=0;i<3;i++) { if(i==0) next.pos=now.pos+1; if(i==1) next.pos=now.pos-1; if(i==2) next.pos=now.pos*2; next.step=now.step+1; if(next.pos==k) return next.step; if(next.pos<0||next.pos>M) continue; if(!visited[next.pos]) { visited[next.pos]=true; Q.push(next); } } } return INF; } int main() { while(cin>>n>>k) { memset(visited,false,sizeof(visited)); if(n<k) cout<<bfs()<<endl; if(n==k) cout<<0<<endl; if(n>k) cout<<n-k<<endl; } }