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?
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 17Sample Output
4Hint
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.
bfs记录每一步的步数,对位置标记,去过的位置不必再去。
代码:
#include <iostream> #include <queue> #include <cstdlib> #include <cstdio> #include <cstring> using namespace std; struct que { int x,t;///x记录位置 t记录步数 }head,temp; int vis[200200]; int main() { ios::sync_with_stdio(false); cin.tie(0); int n,k; while(~scanf("%d%d",&n,&k)) { memset(vis,0,sizeof(vis));//归零 queue<que> q; temp.x = n; vis[n] = 1; temp.t = 0; q.push(temp); while(!q.empty() && q.front().x != k) { head = q.front(); q.pop(); if(head.x < k) { if(!vis[head.x * 2]) { temp.x = head.x * 2; vis[temp.x] = 1; temp.t = head.t + 1; q.push(temp); } if(!vis[head.x + 1]) { temp.x = head.x + 1; vis[temp.x] = 1; temp.t = head.t + 1; q.push(temp); } } if(head.x >= 1 && !vis[head.x - 1]) {///约束条件 temp.x = head.x - 1; vis[temp.x] = 1; temp.t = head.t + 1; q.push(temp); } } printf("%d ",q.front().t); } return 0; }