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catch the cow

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.

思路：bfs 因為要所需時間最短 所以每次一層一層從小的時間點找起來
農夫的移動方向有三種 +1 -1 *2；

#include<iostream> #include<queue> #include<cstdio> #include<cstring> using namespace std;const int maxn=1e5+10; bool visit[maxn]; struct state{int position;int time;state(){}state(int p,int t):position(p),time(t){} };int bfs(int n,int k){queue<state>q;q.push(state(n,0));visit[n]=true;while(!q.empty()){state current=q.front();if(current.position==k){return current.time;}//如果現在的狀態的位置就是牛的位置 返回時間q.pop();//否則將隊首彈出 但是其實值已經保存在currentfor(int i=0;i<3;i++){//三種狀態轉移state next=current;if(i==0) {next.position-=1;} else if(i==1) {next.position+=1;} else{ next.position*=2;} next.time+=1;if(next.position<0||next.position>1e5||visit[next.position]){continue;//位置不合法}q.push(next);visit[next.position]=true;} }}int main(){int n,k;scanf("%d%d",&n,&k);//cin>>n>>k;memset(visit,false,sizeof(visit));printf("%d\n",bfs(n,k));//cout<<bfs(n,k);return 0; }

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