Catch That Cow
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 112355 | Accepted: 35099 |
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.
比较坑的是队列不能定义在函数内部,要定义为全局变量,不然会WA。可能这题对队列容量需求确实比较大,定义为全局变量可以获得更大的队列空间。不过要记得每次清空。
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#include <cstdio> #include <iostream> #include <algorithm> #include <string> #include <cstring> #include <cmath> #include <queue> #include <stack> #include <sstream> #include <map> #define INF 10000 using namespace std; int n,k; int vis[200005]; struct DATA { int v,step; DATA(int vv,int ss) { v=vv; step=ss; } }; typedef struct DATA DATA; queue<DATA> Q; int BFS() { DATA s(n,0); Q.push(s); vis[n]=1; while(Q.size()>0) { DATA cur=Q.front(); Q.pop(); if(cur.v==k) { return cur.step; } int cv=cur.v; int cs=cur.step; if(cv>0 && vis[cv-1]==0) { DATA tmp(cv-1,cs+1); vis[cv-1]=1; Q.push(tmp); } if(cv<=k && vis[cv+1]==0) { DATA tmp(cv+1,cs+1); vis[cv+1]=1; Q.push(tmp); } if(cv<=k && vis[cv*2]==0) { DATA tmp(cv*2,cs+1); vis[cv*2]=1; Q.push(tmp); } } } int main() { while(scanf("%d%d",&n,&k)==2) { memset(vis,0,sizeof(vis)); if(n>=k) { printf("%d\n",n-k); continue; } while(Q.size()>0) Q.pop(); int ans=BFS(); printf("%d\n",ans); } return 0; }