POJ3278(KB1-C 简单搜索)
Catch That Cow
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.
Source
1 //2017-02-22 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <queue> 6 7 using namespace std; 8 9 bool book[200005]; 10 struct node 11 { 12 int pos, step; 13 }; 14 15 int main() 16 { 17 int n, k; 18 while(cin>>n>>k) 19 { 20 memset(book, 0, sizeof(book)); 21 queue<node> q; 22 book[n] = 1; 23 node tmp; 24 tmp.pos = n; 25 tmp.step = 0; 26 q.push(tmp); 27 int pos, step; 28 while(!q.empty()) 29 { 30 pos = q.front().pos; 31 step = q.front().step; 32 q.pop(); 33 if(pos == k){ 34 cout<<step<<endl; 35 break; 36 } 37 if(pos-1==k || pos+1==k || 2*pos==k) 38 { 39 cout<<step+1<<endl; 40 break; 41 } 42 if(pos >= 1 && !book[pos-1]){ 43 tmp.pos = pos-1; 44 tmp.step = step+1; 45 q.push(tmp); 46 book[pos-1] = 1; 47 } 48 if(!book[pos+1]){ 49 book[pos+1] = 1; 50 tmp.pos = pos+1; 51 tmp.step = step+1; 52 q.push(tmp); 53 } 54 if(2*pos<=200005&&!book[pos*2]){ 55 book[2*pos] = 1; 56 tmp.pos = 2*pos; 57 tmp.step = step+1; 58 q.push(tmp); 59 } 60 } 61 } 62 63 return 0; 64 }
