POJ - 3278 Catch That Cow

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.

一维的bfs，每次有三种走法，x+1，x-1，x*2，bfs跑一遍输出步数就行.

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cstring>

using namespace std;

typedef struct zb
{
int x, step;
}node;

int v[999999];

void bfs(int x, int k)
{
node a, b;
queue<node>q;
a.x = x;
a.step = 0;
v[a.x] = 1;
q.push(a);
while(!q.empty())
{
a = q.front();
q.pop();
if(a.x == k)
{
cout<<a.step<<endl;
break;
}
if(a.x + 1 <= k && !v[a.x+1])
{
b.x = a.x+1;
b.step = a.step + 1;
v[b.x] = 1;
q.push(b);
}
if(a.x * 2 <= 200000 && !v[a.x * 2])
{
b.x = a.x * 2;
b.step = a.step + 1;
v[b.x] = 1;
q.push(b);
}
if(a.x - 1 >= 0 && !v[a.x - 1])
{
b.x = a.x-1;
b.step = a.step + 1;
v[b.x] = 1;
q.push(b);
}
}
}

int main()
{
int n, k;
cin>>n>>k;
memset(v,0,sizeof(v));
bfs(n,k);
return 0;
}