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.
#include<stdio.h> #include<string.h> const int maxn = 100001; bool vis[maxn]; int n, k; struct Node { int x, step; }; Node q[maxn]; int bfs() { int i; Node now, next; int head, tail; head = tail = 0; q[tail].x = n;//手动维护一个队列 为啥我用stl的队列会崩 q[tail].step = 0; tail++; vis[n] = true; while (head<tail) { now = q[head];//取队首 head++;//弹出对首 for (i = 0; i<3; i++) { if (i == 0) next.x = now.x - 1; else if (i == 1) next.x = now.x + 1; else next.x = 2 * now.x; if (next.x<0 || next.x >= maxn) continue;//剪枝、排除越界 if (!vis[next.x]) { vis[next.x] = true; next.step = now.step + 1; q[tail].x = next.x; q[tail].step = next.step; tail++; if (next.x == k) return next.step; } } } } int main() { while (scanf("%d%d", &n, &k) != EOF) { memset(vis, false, sizeof(vis)); if (n >= k) printf("%d\n", n - k); else printf("%d\n", bfs()); } return 0; }