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.
一维的bfs，每次有三种走法，x+1，x-1，x*2，bfs跑一遍输出步数就行.
#include <iostream> #include <cstdio> #include <queue> #include <algorithm> #include <cstring> using namespace std; typedef struct zb { int x, step; }node; int v[999999]; void bfs(int x, int k) { node a, b; queue<node>q; a.x = x; a.step = 0; v[a.x] = 1; q.push(a); while(!q.empty()) { a = q.front(); q.pop(); if(a.x == k) { cout<<a.step<<endl; break; } if(a.x + 1 <= k && !v[a.x+1]) { b.x = a.x+1; b.step = a.step + 1; v[b.x] = 1; q.push(b); } if(a.x * 2 <= 200000 && !v[a.x * 2]) { b.x = a.x * 2; b.step = a.step + 1; v[b.x] = 1; q.push(b); } if(a.x - 1 >= 0 && !v[a.x - 1]) { b.x = a.x-1; b.step = a.step + 1; v[b.x] = 1; q.push(b); } } } int main() { int n, k; cin>>n>>k; memset(v,0,sizeof(v)); bfs(n,k); return 0; }