Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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.
ac代码:
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int MAXN = 1e6;
int vis[MAXN], step[MAXN];
queue <int> q;
int main()
{
memset(vis, 0, sizeof(vis));
memset(step, 0, sizeof(step));
int n, k, i, x, dx, flag = 0;
scanf("%d%d", &n, &k);
q.push(n);
vis[n] = 1;
while( !q.empty() )
{
x = q.front();
q.pop();
for( i=0; i<3; i++ )
{
if( 0 == i )
dx = x*2;
else if( 1 == i)
dx = x+1;
else if( 2 == i )
dx = x-1;
if( dx>=0 && dx<MAXN && !vis[dx] )
{
vis[dx] = 1;
step[dx] = step[x]+1;
q.push(dx);
}
if( dx == k )
{
flag = 1;
break;
}
}
if( flag )
break;
}
printf("%d\n", step[k]);
return 0;
}
感冒了,注意力真的分散了,T T
ac代码:
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int MAXN = 1e6;
int vis[MAXN], step[MAXN];
queue <int> q;
int main()
{
memset(vis, 0, sizeof(vis));
memset(step, 0, sizeof(step));
int n, k, i, x, dx, flag = 0;
scanf("%d%d", &n, &k);
q.push(n);
vis[n] = 1;
while( !q.empty() )
{
x = q.front();
q.pop();
for( i=0; i<3; i++ )
{
if( 0 == i )
dx = x*2;
else if( 1 == i)
dx = x+1;
else if( 2 == i )
dx = x-1;
if( dx>=0 && dx<MAXN && !vis[dx] )
{
vis[dx] = 1;
step[dx] = step[x]+1;
q.push(dx);
}
if( dx == k )
{
flag = 1;
break;
}
}
if( flag )
break;
}
printf("%d\n", step[k]);
return 0;
}
感冒了,注意力真的分散了,T T