hdoj_2717Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4193    Accepted Submission(s): 1354

Problem 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

有点记忆化搜索的味道=。=

#include <iostream>#include <cstring>#include <queue>using namespace std;#define MAX 100005bool vis[MAX];typedef struct Point{int x;int cnt;}Point;queue<Point>Q;int bfs(int x, int y){while (!Q.empty()){Q.pop();}Point pre, next;pre.x = x;vis[x] = true;pre.cnt = 0;Q.push(pre);while(!Q.empty()){pre = Q.front();if(pre.x == y){return pre.cnt;}Q.pop();next.x = pre.x + 1;if(next.x <= MAX && !vis[next.x]){vis[next.x] = true;next.cnt = pre.cnt + 1;Q.push(next);}next.x = pre.x - 1;if(next.x >= 0 && !vis[next.x]){vis[next.x] = true;next.cnt = pre.cnt + 1;Q.push(next);}next.x = pre.x * 2;if(next.x <= MAX && !vis[next.x]){vis[next.x] = true;next.cnt = pre.cnt + 1;Q.push(next);}}return 0;}int main(){int n, k;while(cin >> n >> k){memset(vis, false, sizeof(vis));if(n >= k){cout << n - k << endl;}else{cout << bfs(n, k) << endl;}}}