poj 3278 搜尋

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37800		Accepted: 11724

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

一道bfs的大水題，下面是代碼l:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <queue>using namespace std;const int maxn=100005;int step[maxn];bool vis[maxn];queue<int> q;int n,k;int next,head;int bfs(){    q.push(n);    step[n]=0;    vis[n]=1;    while(!q.empty()){         head=q.front();         q.pop();         for(int i=0;i<3;i++){             if(i==0) next=head-1;             else if(i==1) next=head+1;             else next=head*2;             if(next>maxn||next<0) continue;             if(!vis[next]){                  q.push(next);                  vis[next]=1;                  step[next]=step[head]+1;             }             if(next==k)                 return step[next];         }    }    return -1;}int main(){   cin>>n>>k;   memset(vis,0,sizeof(vis));    if(n>=k)//開始寫的時候忘記寫了       cout<<n-k<<endl;    else       printf("%d\n",bfs());       return 0;}