POJ 3278, Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 13066 Accepted: 3968
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.
Source
USACO 2007 Open Silver
//
#include <iostream>
#include <queue>
using namespace std;
int main(int argc, char* argv[])
{
int N, K;
cin >> N >> K;
const int MAXSIZE = 200002;
int line[MAXSIZE];
memset(line, -1, sizeof(line));
queue<int> q;
q.push(N);
line[N] = 0;
while(!q.empty())
{
int stp = q.front();
q.pop();
int next = stp - 1;
if (next >= 0 && line[next] == -1)
{
line[next] = line[stp] + 1;
q.push(next);
if (next==K)
break;
}
next = stp + 1;
if (next < MAXSIZE && line[next] == -1)
{
line[next] = line[stp] + 1;
q.push(next);
if (next==K)
break;
}
next = stp << 1;
if (next < MAXSIZE && line[next] == -1)
{
line[next] = line[stp] + 1;
q.push(next);
if (next==K)
break;
}
}
cout << line[K] << endl;
return 0;
}
