POJ3275——BFS——Catch That Cow

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

大意：从n到k可以两种变化，一是把n翻倍，二是把n-1或n+1，问最小的步数实现

用BFS，广度优先搜索，使用队列。

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,p,countn,k;
struct edge{
    int x;
    int c;
}a[200010];
queue <edge> q;
int visit[200010];
int bfs()
{
    while(!q.empty()){
           p = q.front().x;
           countn = q.front().c;
           q.pop();
           if(p == k)
            return countn;
           if(!visit[p-1] && p > 0){
                visit[p-1] = 1;
                a[p-1].x = p-1;
                a[p-1].c = countn+1;
                q.push(a[p-1]);
           }
           if(p < k){
               if(!visit[p+1]){
                    visit[p+1] = 1;
                    a[p+1].x = p+1;
                    a[p+1].c = countn+1;
                    q.push(a[p+1]);
               }
               if(!visit[p*2]){
                    visit[p*2] = 1;
                    a[p*2].x = p*2;
                    a[p*2].c = countn+1;
                    q.push(a[p*2]);
               }
           }
    }
}
int main()
{
   while(~scanf("%d%d",&n,&k)){
       memset(visit,0,sizeof(visit));
       memset(a,0,sizeof(a));
        while(!q.empty())
        q.pop();
        visit[n] = 1;
       a[n].x = n;
       a[n].c = 0;
       q.push(a[n]);
       printf("%d\n",bfs());
   }
   return 0;
}