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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 KOutput
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.Sample Input
5 17Sample Output
4Hint
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.
由于移动方式只有 +1 -1 *2 因此如果 起点大于终点，直接相减即可
剩下就是标准的BFS模板
1 /* 2 By:OhYee 3 Github:OhYee 4 Email:oyohyee@oyohyee.com 5 Blog:http://www.cnblogs.com/ohyee/ 6 7 かしこいかわいい？ 8 エリーチカ！ 9 要写出来Хорошо的代码哦~ 10 */ 11 12 #include <cstdio> 13 #include <algorithm> 14 #include <cstring> 15 #include <cmath> 16 #include <string> 17 #include <iostream> 18 #include <vector> 19 #include <list> 20 #include <queue> 21 #include <stack> 22 using namespace std; 23 24 //DEBUG MODE 25 #define debug 0 26 27 //循环 28 #define REP(n) for(int o=0;o<n;o++) 29 30 const int maxn = 100005; 31 32 int BFS(int s,int v) { 33 if(s == v) 34 return 0; 35 if(s > v) 36 return s - v; 37 38 queue<int> Q; 39 bool visited[maxn]; 40 memset(visited,false,sizeof(visited)); 41 int dis[maxn]; 42 memset(dis,0,sizeof(dis)); 43 44 Q.push(s); 45 visited[s] = true; 46 while(!Q.empty()) { 47 int th = Q.front(); 48 Q.pop(); 49 50 //达到终点 51 if(th == v) 52 break; 53 54 //拓展节点 55 56 #define push 57 if(next > maxn || next <= 0) 58 continue;59 if(!visited[next]) {60 Q.push(next);61 visited[next] = true;62 dis[next] = dis[th] + 1;63 }64 65 66 int next; 67 next = th + 1; 68 push; 69 next = th - 1; 70 push; 71 next = th * 2; 72 push; 73 } 74 75 if(dis[v]) 76 return dis[v]; 77 else 78 return -1; 79 } 80 81 bool Do() { 82 int s,v; 83 if(scanf("%d%d",&s,&v)==EOF) 84 return false; 85 printf("%d\n",BFS(s,v)); 86 return true; 87 } 88 89 int main() { 90 while(Do()); 91 return 0; 92 }
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原文地址：http://www.cnblogs.com/ohyee/p/5389479.html
