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?
* 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
#include <iostream> #include <cstring> #include <queue> using namespace std; #define MAX 100005 bool 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; } } }