POJ 3278 : 经典BFS

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8341		Accepted: 2476

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.
#include <cstdio> #include <queue> using namespace std; const int MAXN = 100001; int N, K; int vis[MAXN]; int ret[MAXN]; queue<int> q; int BFS(int s, int d) { if (s == d) return 0; q.push(s); int cur; while (!q.empty()) { cur = q.front(); q.pop(); //if (cur == d) break; // vis[cur] = 1; /* 注意BFS的概念，这里的做法是：如果起点不是终点，把起点放入队列。然后搜索起点周围的节点。注意标记为访问的时间很重要， 并不是在POP的时候标记，而是在以POP出节点为中心向四周扫描的时候，将周围的邻居标记。并且考察邻居是否为终点。而不是考察POP 出的那个节点是否为终点。这样做才能保证BFS找到的第一个解是最优解。 */ if (cur + 1 < MAXN && !vis[cur + 1]) { q.push(cur + 1); ret[cur + 1] = ret[cur] + 1; vis[cur + 1] = 1; } if (cur + 1 == d) break; if (cur - 1 >= 0 && !vis[cur - 1]) { q.push(cur - 1); ret[cur - 1] = ret[cur] + 1; vis[cur - 1] = 1; } if (cur - 1 == d) break; if (cur << 1 < MAXN && !vis[cur << 1]) { q.push(cur << 1); ret[cur << 1] = ret[cur] + 1; vis[cur << 1] = 1; } if (cur << 1 == d) break; }; return ret[d]; } int main() { scanf("%d %d", &N, &K); printf("%d/n", BFS(N, K)); return 0; }