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.analysis: pay attention to the " 2*x" it maybe will make the array bounds.
code:
#include<iostream> #include<cstdio> #include<string.h> #include<math.h> #include<string> #include<map> #include<set> #include<queue> #include<vector> #include<algorithm> using namespace std; const int maxn = 100005; const int INF = 1000000; int ans[maxn] = {0}; int main() { int n,k; cin >> n >> k; for(int i=0;i<maxn;i++){ ans[i] = INF; } queue<int> que; que.push(n); ans[n] = 0; while(que.size()){ int temp = que.front(); int md[3] = {1,-1,temp}; que.pop(); for(int i=0;i<3;i++){ int d = temp + md[i]; if(d < maxn){ // attention ! in order to prevent the array from crossing the border. if(ans[d] == INF){ que.push(d); ans[d] = ans[temp] + 1; } } } if(ans[k] != INF){ cout << ans[k] <<endl; break; } } return 0; }
