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.
Source
题解
题意：给定两个整数n和k，通过 n+1或n-1 或n*2 这3种操作，使得n==k，输出最少的操作次数
bfs，找到的第一组就是最小步数
#include<iostream> #include<cstdlib> #include<queue> #define N 1000000 using namespace std; struct node { int x,step; }; queue <node> Q; node first,now; bool vis[N+10]; int n,k,ans; int check(int x) { if (x<0 || x>N ||vis[x]) return 0; return 1; } void extend(node in) { node p = in; p.step ++; p.x = in.x*2; if (check(p.x)) vis[p.x] = true,Q.push(p); p.x = in.x+1; if (check(p.x)) vis[p.x] = true,Q.push(p); p.x = in.x-1; if (check(p.x)) vis[p.x] = true,Q.push(p); } int bfs() { Q.push(first); while (!Q.empty()) { now = Q.front(); Q.pop(); if (now.x == k) return now.step; extend(now); } } int main() { cin>>n>>k; first.x = n; cout<<bfs(); }