问题描述:
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.
解题思路:
bfs,三个方向搜索,x=x+1,x=x-1,x=x*2,最先搜索到的就是用时最短的。
代码:
#include<cstdio> #include<queue> #include<cstring> #include<iostream> using namespace std; int n,k; int a[100010]; struct node { int x; int step; }now,net; int bfs(int x) { queue<node> q; now.x=x; now.step=0; q.push(now); while(q.size()) { now=q.front(); q.pop(); //三种情况 if(now.x==k)return now.step; net.x=now.x+1; if(net.x>=0&&net.x<=100000&&a[net.x]==0) { a[net.x]=1; net.step=now.step+1; q.push(net); } net.x=now.x-1; if(net.x>=0&&net.x<=100000&&a[net.x]==0) { a[net.x]=1; net.step=now.step+1; q.push(net); } net.x=now.x*2; if(net.x>=0&&net.x<=100000&&a[net.x]==0) { a[net.x]=1; net.step=now.step+1; q.push(net); } } return -1; } int main() { while(~scanf("%d%d",&n,&k)) { memset(a,0,sizeof(a)); a[n]=1; int ans=bfs(n); printf("%d\n",ans); } return 0; }
原文地址:https://www.cnblogs.com/boboyuzz/p/10489166.html