Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 61758 | Accepted: 19306 |
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
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题意：
你在一个一维坐标的n位置，牛在k位置，你要从n到k，抓到那头牛。你可以有三种走法，n+1，n-1，或者n*2直接跳。求你抓到那头牛的最短步数。
思路：
简单广搜的思想。状态跳转的时候有三种跳转的方式，将新的状态放到队列中，再不断提取队列中最前面的状态，直到找到k位置。
#include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; int n,k; int visit[100005]; int mark[100005]; int t; struct nodes { int x; int m; }; nodes node1,node2; void bfs() { queue<nodes> q; while(!q.empty()) q.pop(); node1.x=n; node1.m=0; q.push(node1); while(!q.empty()) { node1=q.front(); q.pop(); if(visit[node1.x]==k) { cout<<node1.m<<endl; break; } node2.x=node1.x+1; if(node2.x>=0&&node2.x<=100000&&!mark[node2.x]) { mark[node2.x]=1; node2.m=node1.m+1; q.push(node2); } node2.x=node1.x-1; if(node2.x>=0&&node2.x<=100000&&!mark[node2.x]) { mark[node2.x]=1; node2.m=node1.m+1; q.push(node2); } node2.x=node1.x*2; if(node2.x>=0&&node2.x<=100000&&!mark[node2.x]) { mark[node2.x]=1; node2.m=node1.m+1; q.push(node2); } } } int main() { while(~scanf("%d%d",&n,&k)) { memset(visit,-1,sizeof(visit)); memset(mark,0,sizeof(mark)); visit[k]=k; mark[n]=1; bfs(); } }