题目：
Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
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
分析：题目大意是在数轴上给定任意起点n、终点k，对任意的点坐标x每次有3种走法：x-1，x+1，2*x。每走一次花费时间为1minute，问从n到k最少需要花费多少时间？
该题是最短路径问题，于是可以用BFS搜索，每次往三个方向BFS，直到到达k，每走一步记录当前时间。
//simonPR #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <queue> using namespace std; const int maxn=100005; int n,k,vis[maxn]; struct catche{ int site,times; //记录点和所花的时间。 }; queue<catche> q; void init(); void init(){ while(!q.empty()) q.pop(); memset(vis,0,sizeof(vis)); } int bfs(int n,int k); int bfs(int n,int k){ if(n==k) return 0; catche now,news,start; start.site=n; start.times=0; q.push(start); //将起点入队列。 vis[n]=1; while(!q.empty()){ int ts; now=q.front(); q.pop(); for(int i=0;i<3;i++){ //分三个方向BFS。 if(i==0) ts=now.site-1; else if(i==1) ts=now.site+1; else ts=2*now.site; if(ts>maxn||vis[ts]==1) continue; //如果已经访问过了或超出数据范围则跳过。 if(ts==k) return now.times+1; //到达终点K，返回时间。 if(vis[ts]==0) //更新点信息，并将新点入队列。 { vis[ts]=1; news.site=ts; news.times=now.times+1; q.push(news); } } } } int main() { scanf("%d%d",&n,&k); init(); //初始化。 int ans=bfs(n,k); printf("%d ",ans); return 0; }