Catch That Cow
题目描述
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?
输入
Line 1: Two space-separated integers: N and K
输出
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
示例输入
5 17
示例输出
4
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1 #include<stdio.h> 2 #include<string.h> 3 #include<iostream> 4 #include<queue> 5 using namespace std; 6 7 int vis[100100]; 8 struct node 9 { 10 int date; 11 int step; 12 }; 13 int bfs(int s, int t) 14 { 15 queue<node>que; 16 struct node tmp,ans; 17 int tx; 18 ans.date = s; 19 ans.step = 0; 20 que.push(ans); 21 vis[s] = 1; 22 while(!que.empty()) 23 { 24 tmp = que.front(); 25 que.pop(); 26 tx = tmp.date-1; 27 if(tx < 0 || tx > 100000); 28 else if(!vis[tx]) 29 { 30 vis[tx] = 1; 31 ans.date = tx; 32 ans.step = tmp.step+1; 33 que.push(ans); 34 if(ans.date == t) 35 return ans.step; 36 } 37 tx = tmp.date+1; 38 if(tx < 0 || tx > 100000); 39 else if(!vis[tx]) 40 { 41 vis[tx] = 1; 42 ans.date = tx; 43 ans.step = tmp.step+1; 44 que.push(ans); 45 if(ans.date == t) 46 return ans.step; 47 } 48 tx = tmp.date*2; 49 if(tx < 0 || tx > 100000); 50 else if(!vis[tx]) 51 { 52 vis[tx] = 1; 53 ans.date = tx; 54 ans.step = tmp.step+1; 55 que.push(ans); 56 if(ans.date == t) 57 return ans.step; 58 } 59 } 60 } 61 62 int main() 63 { 64 int s,t; 65 scanf("%d %d",&s,&t); 66 if(s == t) 67 { 68 printf("0\n"); 69 } 70 else 71 { 72 memset(vis,0,sizeof(vis)); 73 printf("%d\n",bfs(s,t)); 74 } 75 return 0; 76 }
