M - Catch That Cow
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.
#include<iostream> #include<cstring> #include<algorithm> #include<stdio.h> #include<queue> typedef long long ll; #define maxn 100010 using namespace std; #define cle(n) memset(n,0,sizeof(n)) int n,k; int vis[2*maxn]; struct Node { int x,step; Node(){} Node(int x,int step):x(x),step(step){} }; int bfs() { queue<Node>que; que.push(Node(n,0)); vis[n]=1; while(que.size()) { Node node=que.front(); que.pop(); for(int i=0;i<3;i++) { int nx=0; if(i==0) nx=node.x-1; if(i==1) nx=node.x+1; if(i==2) nx=node.x*2; if(nx==k) { cout<<node.step+1<<endl; return 0; } if(nx<0||nx>100000) continue; if(!vis[nx]) { vis[nx]=1; que.push(Node(nx,node.step+1)); } } } } int main() { while(~scanf("%d%d",&n,&k)) { cle(vis); if(n<k) bfs(); if(n==k) cout<<0<<endl; if(n>k) cout<<n-k<<endl; } return 0; }
