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
思路:
这个题目又开发广搜的一个新的功能:求出图中两点的最短距离
这个证明是通过搜索树得到的,想象一个点可以扩展为其他三个点,那么被扩展出来的这三个点到最初始点的那个距离就是最短的,因为如果不是最短的,那么这三个点在之前就会出现
其实质就是一个最优路线上的每一个被扩展出的点都是最优的。
#include <iostream> #include <cstring> #include <queue> #define MAX 100007 #define INF 0x7fffffff using namespace std; int n,k; int vis[MAX]; int dis[MAX]; int bfs() { queue<int> q; q.push(n); int next; while(!q.empty()) { int tmp = q.front(); q.pop(); for(int i = 1;i <= 3;i++) { if(i == 1) next = tmp-1; else if(i == 2) next = tmp+1; else next = tmp*2; if(next<0 || next>MAX) continue; if(!vis[next]) { vis[next] = 1; dis[next] = dis[tmp]+1; q.push(next); } if(next == k) return dis[k]; } } } int main() { while(cin>>n>>k) { memset(vis,0,sizeof(vis)); vis[n] = 1; if(n >= k) cout<<n-k<<endl; else cout<<bfs()<<endl; } return 0; }