HDU 2717 Catch That Cow【BFS】

Catch That Cow

Time Limit : 5000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 6   Accepted Submission(s) : 4

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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?

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.

简单的广搜题

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
int vis[1000010];
struct node
{
int x,step;
};
void BFS(int a,int b)
{
queue<node>q;
node cur,next;
cur.x=a;
cur.step=0;
vis[a]=1;
q.push(cur);
while(!q.empty())
{
cur=q.front();
q.pop();

if(cur.x==b)
{
printf("%d\n",cur.step);return;
}
next.x=cur.x-1;
if(next.x>=0&&next.x<=100000&&vis[next.x]!=1)
{
vis[next.x]=1;
next.step=cur.step+1;
q.push(next);
}
next.x=cur.x+1;
if(next.x>=0&&next.x<=100000&&vis[next.x]!=1)
{
vis[next.x]=1;
next.step=cur.step+1;
q.push(next);
}
next.x=cur.x*2;
if(next.x>=0&&next.x<=100000&&vis[next.x]!=1)
{
vis[next.x]=1;
next.step=cur.step+1;
q.push(next);
}
}
}
int main (void)
{
int n,k;
while(~scanf("%d%d",&n,&k))
{
memset(vis,0,sizeof(vis));
BFS(n,k);
}
return 0;
}