BFS宽度优先搜索
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.
1 #include<stdio.h> 2 #include<string.h> 3 #include<queue> 4 using namespace std; 5 #define N 100001 6 bool s[N]; 7 struct node 8 { 9 int x,time; 10 }; 11 bool limit(int x) 12 { 13 if(x>=0&&x<=100000) 14 return true; 15 else return false; 16 } 17 void bfs(int n,int k) 18 { 19 queue<node> q; 20 node a,b; 21 memset(s,false,sizeof(s)); 22 a.x=n;a.time=0; 23 s[n]=true; 24 q.push(a); 25 while(!q.empty()) 26 { 27 a=q.front(); 28 q.pop(); 29 if(a.x==k) 30 { 31 printf("%d\n",a.time); 32 return ; 33 } 34 if(limit(a.x*2)&&!s[a.x*2]) 35 { 36 b.x=a.x*2; 37 b.time=a.time+1; 38 s[a.x*2]=true; 39 q.push(b); 40 } 41 if(limit(a.x-1)&&!s[a.x-1]) 42 { 43 b.x=a.x-1; 44 b.time=a.time+1; 45 s[a.x-1]=true; 46 q.push(b); 47 } 48 if(limit(a.x+1)&&!s[a.x+1]) 49 { 50 b.x=a.x+1; 51 b.time=a.time+1; 52 s[a.x+1]=true; 53 q.push(b); 54 } 55 } 56 } 57 int main() 58 { 59 int n,k; 60 while(scanf("%d%d",&n,&k)!=EOF) 61 { 62 bfs(n,k); 63 } 64 return 0; 65 }
