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 17Sample Output
4Hint
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.
继续整理简单搜索
poj不支持万能头
#include <cstdio> #include <iostream> #include <queue> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; const int inf=0x3f3f3f3f; const int mod=1e9+7; const int N=1e5+30; int n,m,ans; bool vis[N]; int step[N]; int dir[3]= {1,-1,2}; void bfs(int x) { queue<int>q; q.push(x); vis[x]=1; while(q.size()) { int tmp=q.front(); q.pop(); for(int i=0; i<3; i++) { int xx; if(dir[i]!=2) xx=tmp+dir[i]; else xx=tmp*2; if(xx>=0&&xx<N&&!vis[xx]) { q.push(xx); vis[xx]=1; step[xx]=step[tmp]+1; if(xx==m) { ans=step[xx]; return ; } } } } } int main() { while(~scanf("%d%d",&n,&m)) { memset(step,0,sizeof(step)); memset(vis,0,sizeof(vis)); ans=0; bfs(n); cout<<ans<<'\n'; } return 0; }
