Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8999 Accepted Submission(s): 2837
Problem 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
HintThe 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.
睡前一水~ 好久没做过搜索了 注意剪枝&&&&&&&&
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std; struct Node { int x,step; Node(){} Node(int x,int step):x(x),step(step){} }; int n,k; bool vis[199999]; int bfs() { queue
Q; Node cur,next; Q.push(Node(n,0)); while(!Q.empty()) { cur=Q.front(); Q.pop(); for(int i=0;i<3;i++) { if(i==0) next.x=cur.x+1; if(i==1) next.x=cur.x-1; if(i==2) next.x=cur.x*2; next.step=cur.step+1; if(next.x==k) return next.step; if(next.x<0||next.x>100000) continue; if(!vis[next.x]) { vis[next.x]=true; Q.push(next); } } } } int main() { while(cin>>n>>k) { memset(vis,0,sizeof(vis)); if(n
k) cout<