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.
不解释 超简单BFS 思想，值得注意的是有可能输入的N和K相等，此时应该输出0
其他没什么问题了
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 #include <queue> 5 using namespace std; 6 struct node 7 { 8 int x,time; 9 }; 10 int N,K; 11 int flag[100005]; 12 bool judge(node p) 13 { 14 if(p.x<0 || p.x>100000 || p.time>flag[p.x]) 15 { 16 return false; 17 } 18 return true; 19 } 20 void bfs() 21 { 22 node p,temp; 23 int i; 24 flag[N]=0; 25 p.x=N; 26 p.time=0; 27 queue<node>q; 28 while(!q.empty()) 29 { 30 q.pop(); 31 } 32 q.push(p); 33 while(!q.empty()) 34 { 35 p=q.front(); 36 q.pop(); 37 for(i=0;i<3;i++) 38 { 39 temp=p; 40 if(i==0) 41 { 42 temp.x+=1; 43 } 44 if(i==1) 45 { 46 temp.x-=1; 47 } 48 if(i==2) 49 { 50 temp.x*=2; 51 } 52 temp.time=p.time+1; 53 if(!judge(temp)) 54 { 55 continue; 56 } 57 if(temp.x==K) 58 { 59 printf("%d\n",temp.time); 60 return; 61 } 62 flag[temp.x]=temp.time; 63 q.push(temp); 64 } 65 } 66 } 67 int main() 68 { 69 while(~scanf("%d%d",&N,&K)) 70 { 71 if(N==K) 72 { 73 printf("0\n"); 74 continue; 75 } 76 memset(flag,0x7F,sizeof(flag)); 77 bfs(); 78 } 79 return 0; 80 }