poj 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 30980		Accepted: 9556

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.

Source

USACO 2007 Open Silver

 

bfs的一道题目。

#include
#include
#include
#include
#include
#define N 200005

using namespace std;

bool vis[N];
int step[N];
int n,k;
queue q;

void bfs(int n,int k)
{
    q.push(n);
    vis[n]=true;
    step[n]=0;
    while(!q.empty())
    {
        n=q.front();
        q.pop();
        if(!vis[n+1]&&n+1<=100000&&n+1>=0)
        {
            vis[n+1]=true;
            q.push(n+1);
            step[n+1]=step[n]+1;
            if (n==k) return;
        }
        if(!vis[n-1]&&n-1<=100000&&n-1>=0)
        {
            vis[n-1]=true;
            q.push(n-1);
            step[n-1]=step[n]+1;
            if (n==k) return;
        }
        if(!vis[2*n]&&2*n<=100000&&2*n>0)
        {
            vis[2*n]=true;
            q.push(2*n);
            step[2*n]=step[n]+1;
            if (n==k) return;
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        memset(step,0,sizeof(step));
        memset(vis,false,sizeof(vis));
        if(n>=k)
           cout<