http://poj.org/problem?id=3278
Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 146388 | Accepted: 44997 |
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 <bits/stdc++.h> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <algorithm> #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdio.h> #include <queue> #include <stack>; #include <map> #include <set> #include <string.h> #include <vector> #define ME(x , y) memset(x , y , sizeof(x)) #define SF(n) scanf("%d" , &n) #define rep(i , n) for(int i = 0 ; i < n ; i ++) #define INF 0x3f3f3f3f #define mod 1000000007 using namespace std; typedef long long ll ; int vis[1300009]; int ans[1300009]; int n , m ; void bfs(int b) { queue<int>q; memset(vis , 0 , sizeof(vis)); memset(ans , 0 , sizeof(ans)); if(n < m) { q.push(b); vis[b] = 1 ; ans[b] = 0 ; while(!q.empty()) { int x = q.front(); if(x == m) { printf("%d\n" , ans[m]); break ; } q.pop() ; if(x >= 0 && x <= 100000 && !vis[x*2]) { vis[2*x] = 1 ; ans[2*x] += vis[x] + ans[x]; q.push(2*x); } if(x >= 0 && !vis[x+1]) { vis[x+1] = 1 ; ans[x+1] += vis[x] + ans[x]; q.push(x+1); } if(x - 1 >= 0 && !vis[x-1]) { vis[x-1] = 1 ; ans[x-1] += vis[x] + ans[x]; q.push(x-1); } } } else { printf("%d\n" , n - m); } } int main() { while(~scanf("%d%d" , &n , &m)) { bfs(n); } return 0; }