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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 54911 | Accepted: 17176 |
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
Output
Sample Input
5 17
Sample Output
4
Hint
#include<iostream> #include<queue> #include<cstring> using namespace std; #define N 100005 struct Node { int x; int step; }; queue<struct Node>q; bool visit[N]; bool ok(int x) { if(x>=0 && x<=100000) return 1; return 0; } int cal(int i,int x) { if(i==0) return x+1; else if(i==1) return x-1; else return x*2; } void bfs(int a,int b) { memset(visit,0,sizeof(visit)); while(!q.empty()) { q.pop(); } Node aa; aa.step=1; aa.x=a; q.push(aa); visit[aa.x]=1; while(!q.empty()) { Node tmp=q.front(); q.pop(); for(int i=0; i<3; i++) { int y=cal(i,tmp.x); if(ok(y)&&!visit[y]) { if(y==b) cout<<tmp.step<<endl; else { aa.step=tmp.step+1; aa.x=y; q.push(aa); }
visit[y]=1; } } } } int main() { int n,k; cin>>n>>k; if(k>n) bfs(n,k); else cout<<(n-k); return 0; }
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原文地址：http://www.cnblogs.com/jasonlixuetao/p/4531101.html
