模板题之广度搜索 HDU2717 Catch That Cow

Problem 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.
 

 很简单的思路，搜索最短路径，每次的选择都是三种，要找到最短的路径，显然BFS，废话不说，帖代码！

#include<cstdio>#include<queue>#include<iostream>#include<string.h>#define mnum 100001using namespace std;int vis[mnum],time[mnum];int n,k;void bfs(){queue<int>q;q.push(n);while(!q.empty()){int cur=q.front();vis[cur]=1;q.pop();if(cur==k){printf("%d\n",time[cur]);break;}else {int t;int tm[3]={cur+1,cur-1,cur*2};for(int i=0;i<3;i++){t=tm[i];if(t>=0&&t<=mnum-1&&vis[t]==0){time[t]=time[cur]+1;vis[t]=1;  //注意这里vis[t]要设为一，意思是已经入队的地方就不要再入队了（因为最短路与回头路是矛盾的！）q.push(t);}}}}}int main(){while(scanf("%d%d",&n,&k)!=EOF){memset(vis,0,sizeof(vis));memset(time,0,sizeof(time));bfs();}}