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.
题解
bfs基础题 题目大概意思是 在一条横坐标轴上 牛跑了 农民要抓牛 牛不知农民要抓牛 牛不动 农民要去追牛只有三种方式 1.前进一步 2.后退一步 3.任意门传送至自己当前坐标的2倍
目标是求最少步数
eg:题目所给 5 → 17 5 → 10 → 20 → 19 → 18 → 17 or 5 → 10 → 9 → 18 → 17
图示 红色代表目标结点 肉色代表已访问过
BFS伪代码 相当于一个模板 //网上找的 比较便于理解
1 q.push(head); 2 while(!q.empty()){ 3 temp=q.front(); 4 q.pop(); 5 if(temp为目标状态) 输出 6 if(temp不合法) continue; 7 if(temp合法) q.push(temp+△); 8 //表示所有temp在搜索树中的所有子节点 9 }
bfs广搜 利用队列实现 将当前搜索到的状态的每一子状态压入队列中 并进行记录 检查队列是否为空 如果不为空 就将队首元素弹出 并以这个状态位为根节点
进行 bfs 直到整个队列为空为止
AC代码
1 #include<iostream> 2 #include<queue> 3 using namespace std; 4 queue <int> q; 5 int n,k,temp,t,vis[200010],step[200010]; 6 7 void bfs(){ 8 q.push(n); 9 vis[n]=1; 10 while(!q.empty()){ 11 temp=q.front(); 12 q.pop(); 13 for(int i=0;i<3;i++){ 14 if(i==0) t=temp-1; 15 if(i==1) t=temp+1; 16 if(i==2) t=temp*2; 17 if(vis[t]==0&&t<=100000){ //未被访问 注意注意！！！！！！要写上t<=100000 因为在×2中 数组会超限 产生 RE 18 vis[t]=1; 19 q.push(t); 20 step[t]=step[temp]+1; //可以只用一个vis[]数组 vis[t]=vis[temp]+1; 21 if(t==k){ 22 cout<<step[t]<<endl; //如果使用vis[]数组记录步数 需要cout<<vis[t]-1<<endl; 因为之前多＋了 1 因此现在－去 23 } 24 } 25 } 26 } 27 } 28 29 int main(){ 30 cin>>n>>k; 31 if(n>=k) cout<<n-k<<endl; 32 else bfs(); 33 return 0; 34 }
AC代码
1 #include<iostream> 2 #include<queue> 3 using namespace std; 4 5 queue <int> q; 6 7 int n,k,temp,t,vis[200010]; 8 9 void bfs(){ 10 q.push(n); 11 vis[n]=1; 12 while(!q.empty()){ 13 temp=q.front(); 14 q.pop(); 15 int t1=temp-1; 16 int t2=temp+1; 17 int t3=temp*2; 18 if(t1==k){ 19 vis[t1]=vis[temp]+1; 20 cout<<vis[t1]-1<<endl; 21 return; 22 } 23 if(t2==k){ 24 vis[t2]=vis[temp]+1; 25 cout<<vis[t2]-1<<endl; 26 return; 27 } 28 if(t3==k){ 29 vis[t3]=vis[temp]+1; 30 cout<<vis[t3]-1<<endl; 31 return; 32 } 33 else{ 34 if(vis[t1]==0&&t1<=100000){ 35 vis[t1]=1; 36 q.push(t1); 37 vis[t1]=vis[temp]+1; 38 } 39 if(vis[t2]==0&&t2<=100000){ 40 vis[t2]=1; 41 q.push(t2); 42 vis[t2]=vis[temp]+1; 43 } 44 if(vis[t3]==0&&t3<=100000){ 45 vis[t3]=1; 46 q.push(t3); 47 vis[t3]=vis[temp]+1; 48 } 49 } 50 } 51 } 52 53 int main(){ 54 cin>>n>>k; 55 if(n>=k) cout<<n-k<<endl; 56 else bfs(); 57 return 0; 58 }